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L'Hôpital's rule

L'Hôpital's rule (/ˌloʊpiːˈtɑːl/ loh-pee-TAHL) is a mathematical theorem used for evaluating the limit of a quotient of two functions, each of which tends to zero or infinity, by taking each function's derivative.

Applying L’Hôpital’s Rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

\[\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}.\]

If \(\underset{x\to a}{\text{lim}}f(x)={L}_{1}\ \text{and}\ \underset{x\to a}{\text{lim}}g(x)={L}_{2}\ne 0,\) then

\[\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}=\frac{{L}_{1}}{{L}_{2}}.\]

However, what happens if \(\underset{x\to a}{\text{lim}}f(x)=0\) and \(\underset{x\to a}{\text{lim}}g(x)=0?\) We call this one of the indeterminate forms, of type \(\frac{0}{0}.\) This is considered an indeterminate form because we cannot determine the exact behavior of \(\frac{f(x)}{g(x)}\) as \(x\to a\) without further analysis. We have seen examples of this earlier in the text. For example, consider

\[\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}\ \text{and}\ \underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{x}.\]

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

\[\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2}=\underset{x\to 2}{\text{lim}}\frac{(x+2)(x-2)}{x-2}=\underset{x\to 2}{\text{lim}}(x+2)=2+2=4.\]

For \(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{x}\) we were able to show, using a geometric argument, that

\[\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{x}=1.\]

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions \(f\) and \(g\) such that \(\underset{x\to a}{\text{lim}}f(x)=0=\underset{x\to a}{\text{lim}}g(x)\) and such that \({g}^{'}(a)\ne 0\) For \(x\) near \(a,\) we can write

\[f(x)\approx f(a)+{f}^{'}(a)(x-a)\]\[g(x)\approx g(a)+{g}^{'}(a)(x-a).\]\[\frac{f(x)}{g(x)}\approx \frac{f(a)+{f}^{'}(a)(x-a)}{g(a)+{g}^{'}(a)(x-a)}.\]\[\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}=\underset{x\to a}{\text{lim}}\frac{{f}^{'}(x)(x-a)}{{g}^{'}(x)(x-a)}=\underset{x\to a}{\text{lim}}\frac{{f}^{'}(x)}{{g}^{'}(x)}.\]

Condensed: the full section is in OpenStax Calculus Volume 1.

Other Indeterminate Forms

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms \(\frac{0}{0}\) and \(\infty \text{/}\infty .\) However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions \(0\cdot \infty ,\) \(\infty -\infty ,\) \({1}^{\infty },\) \({\infty }^{0},\) and \({0}^{0}\) are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form \(\frac{0}{0}\) or \(\infty \text{/}\infty .\)

Condensed: the full section is in OpenStax Calculus Volume 1.

Growth Rates of Functions

Suppose the functions \(f\) and \(g\) both approach infinity as \(x\to \infty .\) Although the values of both functions become arbitrarily large as the values of \(x\) become sufficiently large, sometimes one function is growing more quickly than the other. For example, \(f(x)={x}^{2}\) and \(g(x)={x}^{3}\) both approach infinity as \(x\to \infty .\) However, as shown in the following table, the values of \({x}^{3}\) are growing much faster than the values of \({x}^{2}.\)

\(x\)	\(10\)	\(100\)	\(1000\)	\(10,000\)
\(f(x)={x}^{2}\)	\(100\)	\(10,000\)	\(1,000,000\)	\(100,000,000\)
\(g(x)={x}^{3}\)	\(1000\)	\(1,000,000\)	\(1,000,000,000\)	\(1,000,000,000,000\)

In fact,

\[\underset{x\to \infty }{\text{lim}}\frac{{x}^{3}}{{x}^{2}}=\underset{x\to \infty }{\text{lim}}x=\infty .\ \text{or, equivalently,}\ \underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{{x}^{3}}=\underset{x\to \infty }{\text{lim}}\frac{1}{x}=0.\]

As a result, we say \({x}^{3}\) is growing more rapidly than \({x}^{2}\) as \(x\to \infty .\) On the other hand, for \(f(x)={x}^{2}\) and \(g(x)=3{x}^{2}+4x+1,\) although the values of \(g(x)\) are always greater than the values of \(f(x)\) for \(x>0,\) each value of \(g(x)\) is roughly three times the corresponding value of \(f(x)\) as \(x\to \infty ,\) as shown in the following table. In fact,

\[\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{3{x}^{2}+4x+1}=\frac{1}{3}.\]

\(x\)	\(10\)	\(100\)	\(1000\)	\(10,000\)
\(f(x)={x}^{2}\)	\(100\)	\(10,000\)	\(1,000,000\)	\(100,000,000\)
\(g(x)=3{x}^{2}+4x+1\)	\(341\)	\(30,401\)	\(3,004,001\)	\(300,040,001\)

In this case, we say that \({x}^{2}\) and \(3{x}^{2}+4x+1\) are growing at the same rate as \(x\to \infty .\)

More generally, suppose \(f\) and \(g\) are two functions that approach infinity as \(x\to \infty .\) We say \(g\) grows more rapidly than \(f\) as \(x\to \infty\) if

\[\underset{x\to \infty }{\text{lim}}\frac{g(x)}{f(x)}=\infty ;\ \text{or, equivalently,}\ \underset{x\to \infty }{\text{lim}}\frac{f(x)}{g(x)}=0.\]

On the other hand, if there exists a constant \(M\ne 0\) such that

\[\underset{x\to \infty }{\text{lim}}\frac{f(x)}{g(x)}=M,\]

we say \(f\) and \(g\) grow at the same rate as \(x\to \infty .\)

\(x\)	\(5\)	\(10\)	\(15\)	\(20\)
\({x}^{3}\)	\(125\)	\(1000\)	\(3375\)	\(8000\)
\({x}^{4}\)	\(625\)	\(10,000\)	\(50,625\)	\(160,000\)
\({e}^{x}\)	\(148\)	\(22,026\)	\(3,269,017\)	\(485,165,195\)

\(x\)	\(10\)	\(100\)	\(1000\)	\(10,000\)
\(\text{ln}(x)\)	\(2.303\)	\(4.605\)	\(6.908\)	\(9.210\)
\(\sqrt[3]{x}\)	\(2.154\)	\(4.642\)	\(10\)	\(21.544\)
\(\sqrt{x}\)	\(3.162\)	\(10\)	\(31.623\)	\(100\)

Condensed: the full section is in OpenStax Calculus Volume 1.

Key Concepts

L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form \(\frac{0}{0}\) or \(\infty \text{/}\infty\) arises.
L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form \(\frac{0}{0}\) or \(\infty \text{/}\infty .\)
The exponential function \({e}^{x}\) grows faster than any power function \({x}^{p},\) \(p>0.\)
The logarithmic function \(\text{ln}\ x\) grows more slowly than any power function \({x}^{p},\) \(p>0.\)

L’Hôpital’s Rule

For the following exercises, evaluate the limit.

For the following exercises, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule.

For the following exercises, evaluate the limits with either L’Hôpital’s rule or previously learned methods.

For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly.

L'Hôpital's rule

L'Hôpital's rule (/ˌloʊpiːˈtɑːl/ loh-pee-TAHL) is a mathematical theorem used for evaluating the limit of a quotient of two functions, each of which tends to zero or infinity, by taking each function's derivative. The rule is named after the 17th-century French mathematician Guillaume de l'Hôpital, who published it in his 1696 textbook after learning it from his tutor, the Swiss mathematician Johann Bernoulli.

For two functions \(f\) and \(g\), under most circumstances the limit of their quotient can be evaluated as the quotient of the limits: \(\lim_{x \to c} f(x)/g(x) = {}\)\(\lim_{x \to c} f(x) \big/ \lim_{x \to c} g(x)\). However, if both limits tend to zero (that is, \(\lim_{x \to c}f(x) = {}\)\(\lim_{x \to c}g(x) = 0\)) or if both tend to infinity, this method cannot be applied because the "indeterminate forms" ⁠\(0/0\)⁠ and ⁠\(\infty/\infty\)⁠ are not well defined. L'Hôpital's rule states that in such cases (assuming a non-vanishing derivative in the denominator),\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},\]where ⁠\(f'\)⁠ and ⁠\(g'\)⁠ are the derivatives of ⁠\(f\)⁠ and ⁠\(g\)⁠.

The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be directly evaluated by continuity.

History

Johann Bernoulli was the original discoverer of this result for the indeterminate form \(0/0\). He sent it to Guillaume de l'Hôpital (also written l'Hospital) in a letter dated July 22, 1694. Previously, during his time in Paris, Bernoulli signed a contract with l'Hôpital, in which he agreed to teach the new calculus to l'Hôpital, to inform him of his own mathematical discoveries, which l'Hôpital might use as he pleased, and to refrain from sharing his notes with others. In return, l'Hôpital paid Bernoulli an annual allowance, which dramatically improved Bernoulli's financial situation. L'Hôpital published this rule in his 1696 book Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes (lit. 'Analysis of the Infinitely Small for the Understanding of Curved Lines'). Its original presentation was geometric and L'Hôpital showcased many of the same examples used by Bernoulli. As the first textbook on the differential calculus to be printed, it dominated much of the eighteenth century and helped popularize new mathematics across Europe. The original result was subsequently extended; these are now collectively referred to as l'Hôpital's rules.

General form

The general form of l'Hôpital's rule covers many cases. Let \(c\) and \(L\) be extended real numbers: real numbers, as well as positive and negative infinity. Let \(I\) be an open interval containing \(c\) (for a two-sided limit) or an open interval with endpoint \(c\) (for a one-sided limit, or a limit at infinity if \(c\) is infinite).

Assumption 1: On \(I\setminus \{c\}\), the real-valued functions \(f\) and \(g\) are differentiable with \(g'(x) \ne 0\).

Assumption 2: \(\lim_{x \to c} \frac{f'(x)}{g'(x)} = L\), a finite or infinite limit.

If either

\[a) \lim_{x \to c}f(x) = \lim_{x \to c}g(x) = 0\]

\[b) \lim_{x \to c} |f(x)| = \lim_{x \to c} |g(x)| = \infty,\]

Condensed: the full section is in Wikipedia.

Necessity of conditions: Counterexamples

All four conditions for l'Hôpital's rule are necessary:

Indeterminacy of form: \(\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0\) or \(\pm \infty\) ;
Differentiability of functions: \(f(x)\) and \(g(x)\) are differentiable on an open interval \(I\) except possibly at the limit point \(c\) in \(I\);
Non-zero derivative of denominator: \(g'(x) \ne 0\) for all \(x\) in \(I\) with \(x \ne c\);
Existence of limit of the quotient of the derivatives: \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) exists.

Where one of the above conditions is not satisfied, l'Hôpital's rule is not valid in general, and its conclusion may be false in certain cases.

1. Form is not indeterminate

The necessity of the first condition can be seen by considering the counterexample where the functions are \(f(x) = x + 1\) and \(g(x) = 2x + 1\) and the limit is \(x \to 1\).

The first condition is not satisfied for this counterexample because \(\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = (1) + 1 = 2 \neq 0\) and \(\lim_{x \to 1} g(x) = \lim_{x \to 1} (2x + 1) = 2(1) + 1 = 3 \neq 0\). This means that the form is not indeterminate.

The second and third conditions are satisfied by \(f(x)\) and \(g(x)\). The fourth condition is also satisfied with

\[\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{(x+1)'}{(2x+1)'} = \lim_{x \to 1} \frac{1}{2} = \frac{1}{2}.\]

But the conclusion fails, since

\[\lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{x+1}{2x+1} = \frac{ \lim_{x \to 1} (x+1) }{ \lim_{x \to 1} (2x+1) } = \frac{2}{3} \neq \frac{1}{2}.\]

2. Differentiability of functions

Differentiability of functions is a requirement because if a function is not differentiable, then the derivative of the function is not guaranteed to exist at each point in \(I\). The fact that \(I\) is an open interval is grandfathered in from the hypothesis of the Cauchy's mean value theorem. The notable exception of the possibility of the functions being not differentiable at \(c\) exists because l'Hôpital's rule only requires the derivative to exist as the function approaches \(c\); the derivative does not need to be taken at \(c\).

For example, let \(f(x) = \begin{cases} \sin x, & x\neq0 \\ 1, & x=0 \end{cases}\), \(g(x)=x\), and \(c = 0\). In this case, \(f(x)\) is not differentiable at \(c\). However, since \(f(x)\) is differentiable everywhere except \(c\), then \(\lim_{x \to c} f'(x)\) still exists. Thus, since \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}\) and \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) exists, l'Hôpital's rule still holds.

3. Derivative of denominator is zero

The necessity of the condition that \(g'(x) \ne 0\) near \(c\) (except possibly at \(c\)) can be seen by the following counterexample due to Otto Stolz. Let \(f(x)=x+\sin x \cos x\) and \(g(x) = f(x) e^{\sin x}.\) Then there is no limit for \(f(x)/g(x)\) as \(x \to \infty.\) However,

\[\begin{aligned} \frac{f'(x)}{g'(x)} &= \frac{2\cos^2 x}{(2 \cos^2 x) e^{\sin x} + (x+\sin x \cos x) e^{\sin x} \cos x} \\ &= \frac{2\cos x}{2 \cos x +x+\sin x \cos x} e^{-\sin x}, \end{aligned}\]

which tends to 0 as \(x \to \infty\), while the original expression before cancellation of \(\cos x\) is undefined at infinitely many points. Further examples of this type were found by Ralph P. Boas Jr.

4. Limit of derivatives does not exist

The requirement that the limit \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) exists is essential; if it does not exist, the original limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) may nevertheless exist. Indeed, as \(x\) approaches \(c\), the functions \(f\) or \(g\) may exhibit many oscillations of small amplitude but steep slope, which do not affect \(\lim_{x \to c} \frac{f(x)}{g(x)}\) but do prevent the convergence of \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\).

For example, if \(f(x)=x+\sin(x)\), \(g(x)=x\) and \(c=\infty\), then

\[\frac{f'(x)}{g'(x)}=\frac{1+\cos(x)}{1},\]

which does not approach a limit since cosine oscillates infinitely between 1 and −1. But the ratio of the original functions does approach a limit, since the amplitude of the oscillations of \(f\) becomes small relative to \(g\):

\[\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \left(\frac{x+\sin(x)}{x} \right) = \lim_{x \to \infty} \left(1+\frac{\sin(x)}{x} \right) = 1+0 = 1.\]

In a case such as this, all that can be concluded is that

\[\liminf_{x \to c} \frac{f'(x)}{g'(x)} \leq \liminf_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f(x)}{g(x)} \leq \limsup_{x \to c} \frac{f'(x)}{g'(x)} ,\]

Condensed: the full section is in Wikipedia.

Examples

In the following computations, each application of l'Hôpital's rule is indicated by the symbol \(\stackrel{\mathrm{H}}{=}\).

Here is a basic example involving the exponential function, which involves the indeterminate form \(\frac{0}{0}\) at \(x = 0\):\[\lim_{x \to 0} \frac{e^x - 1}{x^2+x} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x^2+x)} = \lim_{x \to 0} \frac{e^x}{2x+1} = 1.\]
This is a more elaborate example involving \(\frac{0}{0}\). Applying l'Hôpital's rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying the rule three times:\[\begin{aligned} \lim_{x \to 0}{\frac{2\sin(x)-\sin(2x)}{x-\sin(x)}} & \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 0}{\frac{2\cos(x)-2\cos(2x)}{1-\cos(x)}} \\[4pt] & \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 0}{\frac{-2\sin(x)+4\sin(2x)}{\sin(x)}} \\[4pt] & \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 0}{\frac{-2\cos(x)+8\cos(2x)}{\cos(x)}} ={\frac{-2+8}{1}} =6. \end{aligned}\]
Here is an example involving \(\frac{\infty}{\infty}\): \[\lim_{x \to \infty}x^n\cdot e^{-x} =\lim_{x \to \infty}{\frac{x^n}{e^x}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty}{\frac{nx^{n-1}}{e^x}} =n\cdot \lim_{x \to \infty}{\frac{x^{n-1}}{e^x}}.\] Repeatedly apply l'Hôpital's rule until the exponent is zero (if \(n\) is an integer) or negative (if \(n\) is fractional) to conclude that the limit is zero.
Here is an example involving the indeterminate form \(0 \cdot \infty\) (see below), which is rewritten as the form \(\frac{\infty}{\infty}\):\[\lim_{x \to 0^+}x \ln x =\lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -x = 0.\]
Here is an example involving the mortgage repayment formula and \(\frac{0}{0}\). Let \(P\) be the principal (loan amount), \(r\) the interest rate per period and \(n\) the number of periods. When \(r\) is zero, the repayment amount per period is \(\frac{P}{n}\) (since only principal is being repaid); this is consistent with the formula for non-zero interest rates: \[\lim_{r\to 0} \frac{Pr(1+r)^n}{(1+r)^n-1} \ \stackrel{\mathrm{H}}{=} \ P \lim_{r\to 0} \frac{(1+r)^n+rn(1+r)^{n-1}}{n(1+r)^{n-1}} = \frac{P}{n}.\]
One can also use l'Hôpital's rule to prove the following theorem. If \(f\) is twice-differentiable in a neighborhood of \(x\) and its second derivative is continuous on this neighborhood, then\[\begin{aligned} \lim_{h \to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} &= \lim_{h\to 0} \frac{f'(x+h)-f'(x-h)}{2h} \\[4pt] &= \lim_{h\to 0} \frac{f''(x+h) + f''(x-h)}{2} \\[4pt] &= f''(x). \end{aligned}\]
Sometimes l'Hôpital's rule is invoked in a tricky way: suppose \(f(x) + f'(x)\) converges as \(x \to \infty\) and that \(e^x\cdot f(x)\) converges to positive or negative infinity. Then:\[\lim_{x \to \infty }f(x) = \lim_{x \to \infty} \frac{e^x\cdot f(x)}{e^x} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{e^x\bigl(f(x)+f'(x)\bigr)}{e^x} = \lim_{x \to \infty} \bigl(f(x)+f'(x)\bigr),\]and so, \(\lim_{x \to \infty}f(x)\) exists and \(\lim_{x \to \infty}f'(x) = 0.\) (This result remains true without the added hypothesis that \(e^x\cdot f(x)\) converges to positive or negative infinity, but the justification is then incomplete.)

Complications

Sometimes L'Hôpital's rule does not reduce to an obvious limit in a finite number of steps, unless some intermediate simplifications are applied. Examples include the following:

Two applications can lead to a return to the original expression that was to be evaluated:

\[\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \ \stackrel{\mathrm{H}}{=} \ \cdots .\]This situation can be dealt with by substituting \(y=e^x\) and noting that \(y\) goes to infinity as \(x\) goes to infinity; with this substitution, this problem can be solved with a single application of the rule:

\[\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{y\to \infty} \frac{y+y^{-1}}{y-y^{-1}} \ \stackrel{\mathrm{H}}{=} \ \lim_{y\to \infty} \frac{1-y^{-2}}{1+y^{-2}} = \frac{1}{1} = 1.\]

Alternatively, the numerator and denominator can both be multiplied by \(e^x,\) at which point L'Hôpital's rule can immediately be applied successfully

\[\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} = \lim_{x \to \infty} \frac{e^{2x} + 1}{e^{2x} - 1} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{2e^{2x}}{2e^{2x}} = 1.\]

An arbitrarily large number of applications may never lead to an answer even without repeating:

\[\lim_{x \to \infty} \frac{x^\frac1{2}+x^{-\frac1{2}}}{x^\frac1{2}-x^{-\frac1{2}}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{\frac1{2}x^{-\frac1{2}}-\frac{1}{2}x^{-\frac3{2}}}{\frac1{2}x^{-\frac1{2}}+\frac1{2}x^{-\frac3{2}}} \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to \infty} \frac{-\frac1{4}x^{-\frac3{2}}+\frac3{4}x^{-\frac5{2}}}{-\frac1{4}x^{-\frac3{2}}-\frac3{4}x^{-\frac5{2}}} \ \stackrel{\mathrm{H}}{=} \ \cdots .\]

This situation too can be dealt with by a transformation of variables, in this case \(y = \sqrt{x}\):

Condensed: the full section is in Wikipedia.

Other indeterminate forms

Other indeterminate forms, such as \(1^\infty\), \(0^0\), \(\infty^0\), \(0 \cdot \infty\), and \(\infty - \infty\), can sometimes be evaluated using L'Hôpital's rule. We again indicate applications of L'Hopital's rule by \(\stackrel{\mathrm{H}}{=}\).

For example, to evaluate a limit involving \(\infty - \infty\), convert the difference of two functions to a quotient:

\[\begin{aligned} \lim_{x \to 1} \left(\frac{x}{x-1}-\frac1{\ln x} \right) & = \lim_{x \to 1} \frac{x\cdot\ln x -x+1}{(x-1)\cdot\ln x} \\[6pt] & \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 1} \frac{\ln x}{\frac{x-1}{x}+\ln x} \\[6pt] & = \lim_{x \to 1} \frac{x\cdot\ln x}{x-1+x\cdot\ln x} \\[6pt] & \ \stackrel{\mathrm{H}}{=} \ \lim_{x \to 1} \frac{1+\ln x}{1+1+\ln x} = \frac{1+0}{1+1+0}. \end{aligned}\]

L'Hôpital's rule can be used on indeterminate forms involving exponents by using logarithms to "move the exponent down". Here is an example involving the indeterminate form \(0^0\):

\[\lim_{x \to 0^+\!} x^x = \lim_{x \to 0^+\!} e^{\ln (x^x)} = \lim_{x \to 0^+\!} e^{x \cdot \ln x} = \lim_{x \to 0^+\!} \exp(x \cdot \ln x) = \exp({\lim_{x \to 0^+\!\!} \, x \cdot \ln x}).\]

It is valid to move the limit inside the exponential function because this function is continuous. Now the exponent \(x\) has been "moved down". The limit \(\lim_{x \to 0^+}x\cdot\ln x\) is of the indeterminate form 0 · ∞ dealt with in an example above: L'Hôpital may be used to determine that

\[\lim_{x \to 0^+} x \cdot \ln x = 0.\]

Condensed: the full section is in Wikipedia.

Stolz–Cesàro theorem

The Stolz-Cesàro theorem is a similar result involving limits of sequences, but it uses finite difference operators rather than derivatives.

Geometric interpretation: parametric curve and velocity vector

Consider the parametric curve in the xy-plane with coordinates given by the continuous functions \(g(t)\) and \(f(t)\), the locus of points \((g(t),f(t))\), and suppose \(f(c) = g(c) = 0\). The slope of the tangent to the curve at \((g(c),f(c)) = (0,0)\) is the limit of the ratio \(\frac{f(t)}{g(t)}\) as \(t \to c\). The tangent to the curve at the point \((g(t),f(t))\) is the velocity vector \((g'(t),f'(t))\) with slope \(\frac{f'(t)}{g'(t)}\). L'Hôpital's rule then states that the slope of the curve at the origin (\(t = c\)) is the limit of the tangent slope at points approaching the origin, provided that this is defined.

Special case

The proof of L'Hôpital's rule is simple in the case where \(f\) and \(g\) are continuously differentiable at the point \(c\) and where a finite limit is found after the first round of differentiation. This is only a special case of L'Hôpital's rule, because it only applies to functions satisfying stronger conditions than required by the general rule. However, many common functions have continuous derivatives (e.g. polynomials, sine and cosine, exponential functions), so this special case covers most applications.

Suppose that \(f\) and \(g\) are continuously differentiable at a real number \(c\), that \(f(c) = g(c) = 0\), and that \(g'(c) \neq 0\). Then

\[\begin{aligned} & \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f(x)-0}{g(x)-0} = \lim_{x \to c} \frac{f(x)-f(c)}{g(x)-g(c)} \\[6pt] = {} & \lim_{x \to c} \frac{\left(\frac{f(x)-f(c)}{x-c} \right)}{\left(\frac{g(x)-g(c)}{x-c} \right)} = \frac{\lim_{x \to c} \left(\frac{f(x)-f(c)}{x-c} \right)}{\lim_{x \to c} \left(\frac{g(x)-g(c)}{x-c} \right)}= \frac{f'(c)}{g'(c)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}. \end{aligned}\]

This follows from the difference quotient definition of the derivative. The last equality follows from the continuity of the derivatives at \(c\). The limit in the conclusion is not indeterminate because \(g'(c) \ne 0\).

The proof of a more general version of L'Hôpital's rule is given below.

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Evaluate each of the following limits by applying L’Hôpital’s rule.
1. \(\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\ x}{x}\)
2. \(\underset{x\to 1}{\text{lim}}\frac{\text{sin}(\pi x)}{\text{ln}\ x}\)
3. \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{1\text{/}x}\)
4. \(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}\)
Cevabı göster
1. Since the numerator \(1-\text{cos}\ x\to 0\) and the denominator \(x\to 0,\) we can apply L’Hôpital’s rule to evaluate this limit. We have
  \[\begin{array}{ll}\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\ x}{x} & =\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx}(1-\text{cos}\ x)}{\frac{d}{dx}(x)} \\ & =\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{1} \\ & =\frac{\underset{x\to 0}{\text{lim}}(\text{sin}\ x)}{\underset{x\to 0}{\text{lim}}(1)} \\ & =\frac{0}{1}=0.\end{array}\]
2. As \(x\to 1,\) the numerator \(\text{sin}(\pi x)\to 0\) and the denominator \(\text{ln}(x)\to 0.\) Therefore, we can apply L’Hôpital’s rule. We obtain
  \[\begin{array}{ll}\underset{x\to 1}{\text{lim}}\frac{\text{sin}(\pi x)}{\text{ln}\ x} & =\underset{x\to 1}{\text{lim}}\frac{\pi \ \text{cos}(\pi x)}{1\text{/}x} \\ & =\underset{x\to 1}{\text{lim}}(\pi x)\text{cos}(\pi x) \\ & =(\pi \cdot 1)(-1)=\text{-}\pi .\end{array}\]
3. As \(x\to \infty ,\) the numerator \({e}^{1\text{/}x}-1\to 0\) and the denominator \((\frac{1}{x})\to 0.\) Therefore, we can apply L’Hôpital’s rule. We obtain
  \[\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}(\frac{-1}{{x}^{2}})}{(\frac{-1}{{x}^{2}})}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1.\]
4. As \(x\to 0,\) both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
  \[\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\frac{\text{cos}\ x-1}{2x}.\]
  Since the numerator and denominator of this new quotient both approach zero as \(x\to 0,\) we apply L’Hôpital’s rule again. In doing so, we see that
  \[\underset{x\to 0}{\text{lim}}\frac{\text{cos}\ x-1}{2x}=\underset{x\to 0}{\text{lim}}\frac{\text{-}\text{sin}\ x}{2}=0.\]
  Therefore, we conclude that
  \[\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}=0.\]
Evaluate \(\underset{x\to 0}{\text{lim}}\frac{x}{\text{tan}\ x}.\)

Cevabı göster

\(1\)
Evaluate each of the following limits by applying L’Hôpital’s rule.
1. \(\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}\)
2. \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}\)
Cevabı göster
1. Since \(3x+5\) and \(2x+1\) are first-degree polynomials with positive leading coefficients, \(\underset{x\to \infty }{\text{lim}}(3x+5)=\infty\) and \(\underset{x\to \infty }{\text{lim}}(2x+1)=\infty .\) Therefore, we apply L’Hôpital’s rule and obtain
  \[\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3}{2}=\frac{3}{2}.\]
  Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of \(x\) in the denominator. In doing so, we saw that
  \[\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2+1\text{/}x}=\frac{3}{2}.\]
  L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
2. Here, \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ x=\text{-}\infty\) and \(\underset{x\to {0}^{+}}{\text{lim}}\text{cot}\ x=\infty .\) Therefore, we can apply L’Hôpital’s rule and obtain
  \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{-}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{-}x\ {\text{csc}}^{2}x}.\]
  Now as \(x\to {0}^{+},\) \({\text{csc}}^{2}x\to \infty .\) Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of \(\text{csc}\ x\) to write
  \[\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{-}x\ {\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{-}x}.\]
  Now \(\underset{x\to {0}^{+}}{\text{lim}}{\text{sin}}^{2}x=0\) and \(\underset{x\to {0}^{+}}{\text{lim}}x=0,\) so we apply L’Hôpital’s rule again. We find
  \[\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{-}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{2\ \text{sin}\ x\ \text{cos}\ x}{-1}=\frac{0}{-1}=0.\]
  We conclude that
  \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}=0.\]
Evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{5x}.\)

Cevabı göster

\(0\)
Consider \(\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}.\) Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Cevabı göster

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get
\[\frac{d}{dx}({x}^{2}+5)=2x\]
and
\[\frac{d}{dx}(3x+4)=3.\]
At which point we would conclude erroneously that
\[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\underset{x\to 1}{\text{lim}}\frac{2x}{3}=\frac{2}{3}.\]
However, since \(\underset{x\to 1}{\text{lim}}({x}^{2}+5)=6\) and \(\underset{x\to 1}{\text{lim}}(3x+4)=7,\) we actually have
\[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\frac{6}{7}.\]
We can conclude that
\[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}\ne \underset{x\to 1}{\text{lim}}\frac{\frac{d}{dx}({x}^{2}+5)}{\frac{d}{dx}(3x+4)}.\]
Explain why we cannot apply L’Hôpital’s rule to evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{x}.\) Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{x}\) by other means.

Cevabı göster

\(\underset{x\to {0}^{+}}{\text{lim}}\text{cos}\ x=1.\) Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is \(\infty\)
Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}x\ \text{ln}\ x.\)

Cevabı göster

First, rewrite the function \(x\ \text{ln}\ x\) as a quotient to apply L’Hôpital’s rule. If we write
\[x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}x},\]
we see that \(\text{ln}\ x\to \text{-}\infty\) as \(x\to {0}^{+}\) and \(\frac{1}{x}\to \infty\) as \(x\to {0}^{+}.\) Therefore, we can apply L’Hôpital’s rule and obtain
\[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{1\text{/}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{\frac{d}{dx}(\text{ln}\ x)}{\frac{d}{dx}(1\text{/}x)}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{-1\text{/}{x}^{2}}=\underset{x\to {0}^{+}}{\text{lim}}(\text{-}x)=0.\]
We conclude that
\[\underset{x\to {0}^{+}}{\text{lim}}x\ \text{ln}\ x=0.\]
Evaluate \(\underset{x\to 0}{\text{lim}}x\ \text{cot}\ x.\)

Cevabı göster

\(1\)
Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x}).\)

Cevabı göster

By combining the fractions, we can write the function as a quotient. Since the least common denominator is \({x}^{2}\text{tan}\ x,\) we have
\[\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x}=\frac{(\text{tan}\ x)-{x}^{2}}{{x}^{2}\text{tan}\ x}.\]
As \(x\to {0}^{+},\) the numerator \(\text{tan}\ x-{x}^{2}\to 0\) and the denominator \({x}^{2}\text{tan}\ x\to 0.\) Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have
\[\underset{x\to {0}^{+}}{\text{lim}}\frac{(\text{tan}\ x)-{x}^{2}}{{x}^{2}\text{tan}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{({\text{sec}}^{2}x)-2x}{{x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x}.\]
As \(x\to {0}^{+},\) \(({\text{sec}}^{2}x)-2x\to 1\) and \({x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x\to 0.\) Since the denominator is positive as \(x\) approaches zero from the right, we conclude that
\[\underset{x\to {0}^{+}}{\text{lim}}\frac{({\text{sec}}^{2}x)-2x}{{x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x}=\infty .\]
Therefore,
\[\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x})=\infty .\]
Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{x}-\frac{1}{\text{sin}\ x}).\)

Cevabı göster

\(0\)
Evaluate \(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}.\)

Cevabı göster

Let \(y={x}^{1\text{/}x}.\) Then,
\[\text{ln}y=\text{ln}({x}^{1\text{/}x})=\frac{1}{x}\ \text{ln}\ x=\frac{\text{ln}\ x}{x}.\]
We need to evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{x}.\) Applying L’Hôpital’s rule, we obtain
\[\underset{x\to \infty }{\text{lim}}\text{ln}\ y=\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1\text{/}x}{1}=0.\]
Therefore, \(\underset{x\to \infty }{\text{lim}}\text{ln}\ y=0.\) Since the natural logarithm function is continuous, we conclude that
\[\text{ln}(\underset{x\to \infty }{\text{lim}}y)=0,\]
which leads to
\[\underset{x\to \infty }{\text{lim}}y={e}^{\ln (y)}=\text{lim}{e}^{0}=1.\text{(All limits}x\ge \infty \text{.)}\]
Hence,
\[\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}=1.\]
Evaluate \(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}\text{ln}(x)}.\)

Cevabı göster

\(e\)
Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}.\)

Cevabı göster

Let
\[y={x}^{\text{sin}\ x}.\]
Therefore,
\[\text{ln}\ y=\text{ln}({x}^{\text{sin}\ x})=\text{sin}\ x\ \text{ln}\ x.\]
We now evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x.\) Since \(\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x=0\) and \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ x=\text{-}\infty ,\) we have the indeterminate form \(0\cdot \infty .\) To apply L’Hôpital’s rule, we need to rewrite \(\text{sin}\ x\ \text{ln}\ x\) as a fraction. We could write
\[\text{sin}\ x\ \text{ln}\ x=\frac{\text{sin}\ x}{1\text{/}\text{ln}\ x}\]
or
\[\text{sin}\ x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}\text{sin}\ x}=\frac{\text{ln}\ x}{\text{csc}\ x}.\]
Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain
\[\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{sin}\ x}{1\text{/}\text{ln}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{-1\text{/}(x{(\text{ln}\ x)}^{2})}=\underset{x\to {0}^{+}}{\text{lim}}(\text{-}x{(\text{ln}\ x)}^{2}\text{cos}\ x).\]
Unfortunately, we not only have another expression involving the indeterminate form \(0\cdot \infty ,\) but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing
\[\text{sin}\ x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}\text{sin}\ x}=\frac{\text{ln}\ x}{\text{csc}\ x},\]
and applying L’Hôpital’s rule, we obtain
\[\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{csc}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{-}\text{csc}\ x\ \text{cot}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{-1}{x\ \text{csc}\ x\ \text{cot}\ x}.\]
Using the fact that \(\text{csc}\ x=\frac{1}{\text{sin}\ x}\) and \(\text{cot}\ x=\frac{\text{cos}\ x}{\text{sin}\ x},\) we can rewrite the expression on the right-hand side as
\[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{-}{\text{sin}}^{2}x}{x\ \text{cos}\ x}=\underset{x\to {0}^{+}}{\text{lim}}[\frac{\text{sin}\ x}{x}\cdot (\text{-}\text{tan}\ x)]=(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{sin}\ x}{x})\cdot (\underset{x\to {0}^{+}}{\text{lim}}(\text{-}\text{tan}\ x))=1\cdot 0=0.\]
We conclude that \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ y=0.\) Therefore, \(\text{ln}(\underset{x\to {0}^{+}}{\text{lim}}y)=0\) and we have
\[\underset{x\to {0}^{+}}{\text{lim}}y=\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}={e}^{0}=1.\]
Hence,
\[\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}=1.\]
Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}{x}^{x}.\)

Cevabı göster

\(1\)
For each of the following pairs of functions, use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}(\frac{f(x)}{g(x)}).\)
1. \(f(x)={x}^{2}\ \text{and}\ g(x)={e}^{x}\)
2. \(f(x)=\text{ln}(x)\ \text{and}\ g(x)={x}^{2}\)
Cevabı göster
1. Since \(\underset{x\to \infty }{\text{lim}}{x}^{2}=\infty\) and \(\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty ,\) we can use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}[\frac{{x}^{2}}{{e}^{x}}].\) We obtain
  \[\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\frac{2x}{{e}^{x}}.\]
  Since \(\underset{x\to \infty }{\text{lim}}2x=\infty\) and \(\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty ,\) we can apply L’Hôpital’s rule again. Since
  \[\underset{x\to \infty }{\text{lim}}\frac{2x}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\frac{2}{{e}^{x}}=0,\]
  we conclude that
  \[\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{{e}^{x}}=0.\]
  Therefore, \({e}^{x}\) grows more rapidly than \({x}^{2}\) as \(x\to \infty\) (See and ).
  
  \(x\) \(5\) \(10\) \(15\) \(20\)
  \({x}^{2}\) \(25\) \(100\) \(225\) \(400\)
  \({e}^{x}\) \(148\) \(22,026\) \(3,269,017\) \(485,165,195\)
2. Since \(\underset{x\to \infty }{\text{lim}}\text{ln}\ x=\infty\) and \(\underset{x\to \infty }{\text{lim}}{x}^{2}=\infty ,\) we can use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{2}}.\) We obtain
  \[\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{2}}=\underset{x\to \infty }{\text{lim}}\frac{1\text{/}x}{2x}=\underset{x\to \infty }{\text{lim}}\frac{1}{2{x}^{2}}=0.\]
  Thus, \({x}^{2}\) grows more rapidly than \(\text{ln}\ x\) as \(x\to \infty\) (see and ).
  
  \(x\) \(10\) \(100\) \(1000\) \(10,000\)
  \(\text{ln}(x)\) \(2.303\) \(4.605\) \(6.908\) \(9.210\)
  \({x}^{2}\) \(100\) \(10,000\) \(1,000,000\) \(100,000,000\)
Compare the growth rates of \({x}^{100}\) and \({2}^{x}.\)

Cevabı göster

The function \({2}^{x}\) grows faster than \({x}^{100}.\)
Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{x}.\)
Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{{x}^{k}}.\)

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\(\infty\)
Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{k}}.\)
Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{2}-{a}^{2}},\ a\ne 0\).

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\(\frac{1}{2a}\)
Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{3}-{a}^{3}},\ a\ne 0\).
Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{n}-{a}^{n}},\ a\ne 0\).

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\(\frac{1}{n{a}^{n-1}}\)
\(\underset{x\to {0}^{+}}{\text{lim}}{x}^{2}\text{ln}\ x\)
\(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}\)

Cevabı göster

Cannot apply directly; use logarithms
\(\underset{x\to 0}{\text{lim}}{x}^{2\text{/}x}\)
\(\underset{x\to 0}{\text{lim}}\frac{{x}^{2}}{1\text{/}x}\)

Cevabı göster

Cannot apply directly; rewrite as \(\underset{x\to 0}{\text{lim}}{x}^{3}\)
\(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{x}\)
\(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x-3}\)

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\(6\)
\(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x+3}\)
\(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{-2}-1}{x}\)

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\(-2\)
\(\underset{x\to \pi \text{/}2}{\text{lim}}\frac{\text{cos}\ x}{\frac{\pi }{2}-x}\)
\(\underset{x\to \pi }{\text{lim}}\frac{x-\pi }{\text{sin}\ x}\)

Cevabı göster

\(-1\)
\(\underset{x\to 1}{\text{lim}}\frac{x-1}{\text{sin}\ x}\)
\(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{n}-1}{x}\)

Cevabı göster

\(n\)
\(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{n}-1-nx}{{x}^{2}}\)
\(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-\text{tan}\ x}{{x}^{3}}\)

Cevabı göster

\(-\frac{1}{2}\)
\(\underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)
\(\underset{x\to 0}{\text{lim}}\frac{{e}^{x}-x-1}{{x}^{2}}\)

Cevabı göster

\(\frac{1}{2}\)
\(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{tan}\ x}{\sqrt{x}}\)
\(\underset{x\to 1}{\text{lim}}\frac{x-1}{\text{ln}\ x}\)

Cevabı göster

\(1\)

Kendi çalışmalarınızı saklayın

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Kayıt Ol Giriş

Burada kullanılan sembolleri

Tüm tanım, resim ve her harfin anlamı için herhangi bir sembolü tıklayın.

Sık sorulan sorular

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), Wikipedia (CC BY-SA 4.0). Burada tekrar özetlenmiş ve açıklanmıştır. Hatalar bizim. Kaynağı ticari olmayan kullanım için lisanslanmış olduğu için, bu ders ücretsizdir, hiçbir şey satmaz ve aynı lisans altında paylaşılır.

More in Calculus

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