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The Limit Laws

Recognize the basic limit laws.

Evaluating Limits with the Limit Laws

The first two limit laws were stated in and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Example

Try it.

Evaluate each of the following limits using .

\(\underset{x\to 2}{\text{lim}}x\)
\(\underset{x\to 2}{\text{lim}}5\)

Solution

The limit of x as x approaches a is a: \(\underset{x\to 2}{\text{lim}}x=2.\)
The limit of a constant is that constant: \(\underset{x\to 2}{\text{lim}}5=5.\)

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

Limit Laws

Let \(f(x)\) and \(g(x)\) be defined for all \(x\ne a\) over some open interval containing a. Assume that L and M are real numbers such that \(\underset{x\to a}{\text{lim}}f(x)=L\) and \(\underset{x\to a}{\text{lim}}g(x)=M.\) Let c be a constant. Then, each of the following statements holds:

Sum law for limits: \(\underset{x\to a}{\text{lim}}(f(x)+g(x))=\underset{x\to a}{\text{lim}}f(x)+\underset{x\to a}{\text{lim}}g(x)=L+M\)

Difference law for limits: \(\underset{x\to a}{\text{lim}}(f(x)-g(x))=\underset{x\to a}{\text{lim}}f(x)-\underset{x\to a}{\text{lim}}g(x)=L-M\)

Constant multiple law for limits: \(\underset{x\to a}{\text{lim}}cf(x)=c\cdot \underset{x\to a}{\text{lim}}f(x)=cL\)

Product law for limits: \(\underset{x\to a}{\text{lim}}(f(x)\cdot g(x))=\underset{x\to a}{\text{lim}}f(x)\cdot \underset{x\to a}{\text{lim}}g(x)=L\cdot M\)

Quotient law for limits: \(\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}=\frac{L}{M}\) for \(M\ne 0\)

Power law for limits: \(\underset{x\to a}{\text{lim}}{(f(x))}^{n}={(\underset{x\to a}{\text{lim}}f(x))}^{n}={L}^{n}\) for every positive integer n.

Root law for limits: \(\underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\text{lim}}f(x)}=\sqrt[n]{L}\) for all L if n is odd and for \(L\ge 0\) if n is even and \(f(x)\ge 0\).

We now practice applying these limit laws to evaluate a limit.

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

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that \(\underset{x\to a}{\text{lim}}f(x)=f(a).\) This is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for which the rational function is defined.

To see that this theorem holds, consider the polynomial \(p(x)={c}_{n}{x}^{n}+{c}_{n-1}{x}^{n-1}+\cdots +{c}_{1}x+{c}_{0}.\) By applying the sum, constant multiple, and power laws, we end up with

\[\begin{array}{ll}\underset{x\to a}{\text{lim}}p(x) & =\underset{x\to a}{\text{lim}}({c}_{n}{x}^{n}+{c}_{n-1}{x}^{n-1}+\cdots +{c}_{1}x+{c}_{0}) \\ & ={c}_{n}{(\underset{x\to a}{\text{lim}}x)}^{n}+{c}_{n-1}{(\underset{x\to a}{\text{lim}}x)}^{n-1}+\cdots +{c}_{1}(\underset{x\to a}{\text{lim}}x)+\underset{x\to a}{\text{lim}}{c}_{0} \\ & ={c}_{n}{a}^{n}+{c}_{n-1}{a}^{n-1}+\cdots +{c}_{1}a+{c}_{0} \\ & =p(a).\end{array}\]

It now follows from the quotient law that if \(p(x)\) and \(q(x)\) are polynomials for which \(q(a)\ne 0,\) then

\[\underset{x\to a}{\text{lim}}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}.\]

applies this result.

Example

Try it.

Evaluate the \(\underset{x\to 3}{\text{lim}}\frac{2{x}^{2}-3x+1}{5x+4}.\)

Solution

Since 3 is in the domain of the rational function \(f(x)=\frac{2{x}^{2}-3x+1}{5x+4},\) we can calculate the limit by substituting 3 for x into the function. Thus,

\[\underset{x\to 3}{\text{lim}}\frac{2{x}^{2}-3x+1}{5x+4}=\frac{10}{19}.\]

Additional Limit Evaluation Techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for \(\underset{x\to a}{\text{lim}}f(x)\) to exist when \(f(a)\) is undefined. The following observation allows us to evaluate many limits of this type:

If for all \(x\ne a,f(x)=g(x)\) over some open interval containing a, then \(\underset{x\to a}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}g(x).\)

To understand this idea better, consider the limit \(\underset{x\to 1}{\text{lim}}\frac{{x}^{2}-1}{x-1}.\)

The function

\[\begin{array}{ll}f(x) & =\frac{{x}^{2}-1}{x-1} \\ & =\frac{(x-1)(x+1)}{x-1}\end{array}\]

and the function \(g(x)=x+1\) are identical for all values of \(x\ne 1.\) The graphs of these two functions are shown in .

We see that

\[\begin{array}{ll}\underset{x\to 1}{\text{lim}}\frac{{x}^{2}-1}{x-1} & =\underset{x\to 1}{\text{lim}}\frac{(x-1)(x+1)}{x-1} \\ & =\underset{x\to 1}{\text{lim}}(x+1) \\ & =2.\end{array}\]

The limit has the form \(\underset{x\to a}{\text{lim}}\frac{f(x)}{g(x)},\) where \(\underset{x\to a}{\text{lim}}f(x)=0\) and \(\underset{x\to a}{\text{lim}}g(x)=0.\) (In this case, we say that \(f(x)\text{/}g(x)\) has the indeterminate form \(0\text{/}0\text{.)}\) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

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

The Squeeze Theorem

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. illustrates this idea.

Example

Try it.

Apply the squeeze theorem to evaluate \(\underset{x\to 0}{\text{lim}}x\ \text{cos}\ x.\)

Solution

Because \(-1\le \text{cos}\ x\le 1\) for all x, we have \(-|x|\le x\ \text{cos}\ x\le |x|\). Since \(\underset{x\to 0}{\text{lim}}(-|x|)=0=\underset{x\to 0}{\text{lim}}|x|,\) from the squeeze theorem, we obtain \(\underset{x\to 0}{\text{lim}}x\ \text{cos}\ x=0.\) The graphs of \(f(x)=-|x|,g(x)=x\ \text{cos}\ x,\) and \(h(x)=|x|\) are shown in .

We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is \(\underset{\theta \to 0}{\text{lim}}\text{sin}\ \theta .\) Consider the unit circle shown in . In the figure, we see that \(\text{sin}\ \theta\) is the y-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for \(0<\theta <\frac{\pi }{2},0<\text{sin}\ \theta <\theta .\)

Because \(\underset{\theta \to {0}^{+}}{\text{lim}}0=0\) and \(\underset{\theta \to {0}^{+}}{\text{lim}}\theta =0,\) by using the squeeze theorem we conclude that

\[\underset{\theta \to {0}^{+}}{\text{lim}}\text{sin}\ \theta =0.\]

To see that \(\underset{\theta \to {0}^{-}}{\text{lim}}\text{sin}\ \theta =0\) as well, observe that for \(-\frac{\pi }{2}<\theta <0,0<\text{-}\theta <\frac{\pi }{2}\) and hence, \(0<\text{sin}\ (-\theta )<\text{-}\theta .\) Consequently, \(0<-\text{sin}\ \theta <\text{-}\theta .\) It follows that \(0>\text{sin}\ \theta >\theta .\) An application of the squeeze theorem produces the desired limit. Thus, since \(\underset{\theta \to {0}^{+}}{\text{lim}}\text{sin}\ \theta =0\) and \(\underset{\theta \to {0}^{-}}{\text{lim}}\text{sin}\ \theta =0,\)

\[\underset{\theta \to 0}{\text{lim}}\ \text{sin}\ \theta =0.\]

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

Key Concepts

The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each time.
For polynomials and rational functions, \(\underset{x\to a}{\text{lim}}f(x)=f(a).\)
You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying a complex fraction.
The squeeze theorem allows you to find the limit of a function if the function is always greater than one function and less than another function with limits that are known.

Key Equations

Basic Limit Results	\(\underset{x\to a}{\text{lim}}x=a\ \underset{x\to a}{\text{lim}}c=c\)
Important Limits	\(\underset{\theta \to 0}{\text{lim}}\text{sin}\ \theta =0\) \(\underset{\theta \to 0}{\text{lim}}\text{cos}\ \theta =1\) \(\underset{\theta \to 0}{\text{lim}}\frac{\text{sin}\ \theta }{\theta }=1\) \(\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\theta }=0\)

The Limit Laws

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

In the following exercises, use direct substitution to evaluate each limit.

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form \(0\text{/}0.\) Then, evaluate the limit.

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of to simplify the function to help determine the limit.

In the following exercises, assume that \(\underset{x\to 6}{\text{lim}}f(x)=4,\underset{x\to 6}{\text{lim}}g(x)=9,\) and \(\underset{x\to 6}{\text{lim}}h(x)=6.\) Use these three facts and the limit laws to evaluate each limit.

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

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

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate each of the following limits using .
1. \(\underset{x\to 2}{\text{lim}}x\)
2. \(\underset{x\to 2}{\text{lim}}5\)
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1. The limit of x as x approaches a is a: \(\underset{x\to 2}{\text{lim}}x=2.\)
2. The limit of a constant is that constant: \(\underset{x\to 2}{\text{lim}}5=5.\)
Use the limit laws to evaluate \(\underset{x\to -3}{\text{lim}}(4x+2).\)

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Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.
\(\begin{array}{lllll}\underset{x\to -3}{\text{lim}}(4x+2) & =\underset{x\to -3}{\text{lim}}4x+\underset{x\to -3}{\text{lim}}2 & & & \text{Apply the sum law.} \\ & =4\cdot \underset{x\to -3}{\text{lim}}x+\underset{x\to -3}{\text{lim}}2 & & & \text{Apply the constant multiple law.} \\ & =4\cdot (-3)+2=-10. & & & \text{Apply the basic limit results and simplify.}\end{array}\)
Use the limit laws to evaluate \(\underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x+1}{{x}^{3}+4}.\)

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To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.
\(\begin{array}{lllll} \\ \\ \underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x+1}{{x}^{3}+4} & =\frac{\underset{x\to 2}{\text{lim}}(2{x}^{2}-3x+1)}{\underset{x\to 2}{\text{lim}}({x}^{3}+4)} & & & \text{Apply the quotient law, making sure that.}\ {(2)}^{3}+4\ne 0 \\ & =\frac{2\cdot \underset{x\to 2}{\text{lim}}{x}^{2}-3\cdot \underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{\underset{x\to 2}{\text{lim}}{x}^{3}+\underset{x\to 2}{\text{lim}}4} & & & \text{Apply the sum law and constant multiple law.} \\ & =\frac{2\cdot {(\underset{x\to 2}{\text{lim}}x)}^{2}-3\cdot \underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{{(\underset{x\to 2}{\text{lim}}x)}^{3}+\underset{x\to 2}{\text{lim}}4} & & & \text{Apply the power law.} \\ & =\frac{2(4)-3(2)+1}{{(2)}^{3}+4}=\frac{1}{4}. & & & \text{Apply the basic limit laws and simplify.}\end{array}\)
Use the limit laws to evaluate \(\underset{x\to 6}{\text{lim}}(2x-1)\sqrt{x+4}.\) In each step, indicate the limit law applied.

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\(11\sqrt{10}\)
Evaluate the \(\underset{x\to 3}{\text{lim}}\frac{2{x}^{2}-3x+1}{5x+4}.\)

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Since 3 is in the domain of the rational function \(f(x)=\frac{2{x}^{2}-3x+1}{5x+4},\) we can calculate the limit by substituting 3 for x into the function. Thus,
\[\underset{x\to 3}{\text{lim}}\frac{2{x}^{2}-3x+1}{5x+4}=\frac{10}{19}.\]
Evaluate \(\underset{x\to -2}{\text{lim}}(3{x}^{3}-2x+7).\)

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−13;
Evaluate \(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-3x}{2{x}^{2}-5x-3}.\)

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Step 1. The function \(f(x)=\frac{{x}^{2}-3x}{2{x}^{2}-5x-3}\) is undefined for \(x=3.\) In fact, if we substitute 3 into the function we get \(0\text{/}0,\) which is indeterminate. Factoring and canceling is a good strategy:
\[\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-3x}{2{x}^{2}-5x-3}=\underset{x\to 3}{\text{lim}}\frac{x(x-3)}{(x-3)(2x+1)}\]
Step 2. For all \(x\ne 3,\frac{{x}^{2}-3x}{2{x}^{2}-5x-3}=\frac{x}{2x+1}.\) Therefore,
\[\underset{x\to 3}{\text{lim}}\frac{x(x-3)}{(x-3)(2x+1)}=\underset{x\to 3}{\text{lim}}\frac{x}{2x+1}.\]
Step 3. Evaluate using the limit laws:
\[\underset{x\to 3}{\text{lim}}\frac{x}{2x+1}=\frac{3}{7}.\]
Evaluate \(\underset{x\to -3}{\text{lim}}\frac{{x}^{2}+4x+3}{{x}^{2}-9}.\)

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\(\frac{1}{3}\)
Evaluate \(\underset{x\to -1}{\text{lim}}\frac{\sqrt{x+2}-1}{x+1}.\)

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Step 1. \(\frac{\sqrt{x+2}-1}{x+1}\) has the form \(0\text{/}0\) at −1. Let’s begin by multiplying by \(\sqrt{x+2}+1,\) the conjugate of \(\sqrt{x+2}-1,\) on the numerator and denominator:
\[\underset{x\to -1}{\text{lim}}\frac{\sqrt{x+2}-1}{x+1}=\underset{x\to -1}{\text{lim}}\frac{\sqrt{x+2}-1}{x+1}\cdot \frac{\sqrt{x+2}+1}{\sqrt{x+2}+1}.\]
Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the \((x+1)\) in the denominator cancels out in the end:
\[=\underset{x\to -1}{\text{lim}}\frac{x+1}{(x+1)(\sqrt{x+2}+1)}.\]
Step 3. Then we cancel:
\[=\underset{x\to -1}{\text{lim}}\frac{1}{\sqrt{x+2}+1}.\]
Step 4. Last, we apply the limit laws:
\[\underset{x\to -1}{\text{lim}}\frac{1}{\sqrt{x+2}+1}=\frac{1}{2}.\]
Evaluate \(\underset{x\to 5}{\text{lim}}\frac{\sqrt{x-1}-2}{x-5}.\)

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\(\frac{1}{4}\)
Evaluate \(\underset{x\to 1}{\text{lim}}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}.\)

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Step 1. \(\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\) has the form \(0\text{/}0\) at 1. We simplify the algebraic fraction by multiplying by \(2(x+1)\text{/}2(x+1):\)
\[\underset{x\to 1}{\text{lim}}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}=\underset{x\to 1}{\text{lim}}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}\cdot \frac{2(x+1)}{2(x+1)}.\]
Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor \((x-1)\text{:}\)
\[=\underset{x\to 1}{\text{lim}}\frac{2-(x+1)}{2(x-1)(x+1)}.\]
Step 3. Then, we simplify the numerator:
\[=\underset{x\to 1}{\text{lim}}\frac{-x+1}{2(x-1)(x+1)}.\]
Step 4. Now we factor out −1 from the numerator:
\[=\underset{x\to 1}{\text{lim}}\frac{-(x-1)}{2(x-1)(x+1)}.\]
Step 5. Then, we cancel the common factors of \((x-1)\text{:}\)
\[=\underset{x\to 1}{\text{lim}}\frac{-1}{2(x+1)}.\]
Step 6. Last, we evaluate using the limit laws:
\[\underset{x\to 1}{\text{lim}}\frac{-1}{2(x+1)}=-\frac{1}{4}.\]
Evaluate \(\underset{x\to -3}{\text{lim}}\frac{\frac{1}{x+2}+1}{x+3}.\)

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−1;
Evaluate \(\underset{x\to 0}{\text{lim}}(\frac{1}{x}+\frac{5}{x(x-5)}).\)

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Both \(1\text{/}x\) and \(5\text{/}x(x-5)\) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that
\[\ \begin{array}{ll}\frac{1}{x}+\frac{5}{x(x-5)} & =\frac{x-5+5}{x(x-5)} \\ & =\frac{x}{x(x-5)}.\end{array}\]
Thus,
\[\begin{array}{ll}\underset{x\to 0}{\text{lim}}(\frac{1}{x}+\frac{5}{x(x-5)}) & =\underset{x\to 0}{\text{lim}}\frac{x}{x(x-5)} \\ & =\underset{x\to 0}{\text{lim}}\frac{1}{x-5} \\ & =-\frac{1}{5}.\end{array}\]
Evaluate \(\underset{x\to 3}{\text{lim}}(\frac{1}{x-3}-\frac{4}{{x}^{2}-2x-3}).\)

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\(\frac{1}{4}\)
Evaluate each of the following limits, if possible.
1. \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}\)
2. \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}\)
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illustrates the function \(f(x)=\sqrt{x-3}\) and aids in our understanding of these limits.
1. The function \(f(x)=\sqrt{x-3}\) is defined over the interval \([3,\text{+}\infty ).\) Since this function is not defined to the left of 3, we cannot apply the limit laws to compute \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}.\) In fact, since \(f(x)=\sqrt{x-3}\) is undefined to the left of 3, \(\underset{x\to {3}^{-}}{\text{lim}}\sqrt{x-3}\) does not exist.
2. Since \(f(x)=\sqrt{x-3}\) is defined to the right of 3, the limit laws do apply to \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}.\) By applying these limit laws we obtain \(\underset{x\to {3}^{+}}{\text{lim}}\sqrt{x-3}=0.\)
For \(f(x)=\{\begin{array}{ll}4x-3 & \text{if}\ x<2 \\ {(x-3)}^{2} & \text{if}\ x\ge 2\end{array},\) evaluate each of the following limits:
1. \(\underset{x\to {2}^{-}}{\text{lim}}f(x)\)
2. \(\underset{x\to {2}^{+}}{\text{lim}}f(x)\)
3. \(\underset{x\to 2}{\text{lim}}f(x)\)
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illustrates the function \(f(x)\) and aids in our understanding of these limits.
1. Since \(f(x)=4x-3\) for all x in \((\text{-}\infty ,2),\) replace \(f(x)\) in the limit with \(4x-3\) and apply the limit laws:
  \[\underset{x\to {2}^{-}}{\text{lim}}f(x)=\underset{x\to {2}^{-}}{\text{lim}}(4x-3)=5.\]
2. Since \(f(x)={(x-3)}^{2}\) for all x in \((2,\text{+}\infty ),\) replace \(f(x)\) in the limit with \({(x-3)}^{2}\) and apply the limit laws:
  \[\underset{x\to {2}^{+}}{\text{lim}}f(x)=\underset{x\to {2}^{+}}{\text{lim}}{(x-3)}^{2}=1.\]
3. Since \(\underset{x\to {2}^{-}}{\text{lim}}f(x)=5\) and \(\underset{x\to {2}^{+}}{\text{lim}}f(x)=1,\) we conclude that \(\underset{x\to 2}{\text{lim}}f(x)\) does not exist.
Graph \(f(x)=\{\begin{array}{l}-x-2\ \text{if}\ x<\text{-}1 \\ 2\ \text{if}\ x=-1 \\ {x}^{3}\ \text{if}\ x>\text{-}1\end{array}\) and evaluate \(\underset{x\to {-1}^{-}}{\text{lim}}f(x).\)

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\(\underset{x\to {-1}^{-}}{\text{lim}}f(x)=-1\)
Evaluate \(\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{{x}^{2}-2x}.\)

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Step 1. After substituting in \(x=2,\) we see that this limit has the form \(-1\text{/}0.\) That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of \(\frac{x-3}{x(x-2)}\) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
\[\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{{x}^{2}-2x}=\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{x(x-2)}.\]
Step 2. Since \(x-2\) is the only part of the denominator that is zero when 2 is substituted, we then separate \(1\text{/}(x-2)\) from the rest of the function:
\[=\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{x}\cdot \frac{1}{x-2}.\]
Step 3. \(\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{x}=-\frac{1}{2}\) and \(\underset{x\to {2}^{-}}{\text{lim}}\frac{1}{x-2}=\text{-}\infty .\) Therefore, the product of \((x-3)\text{/}x\) and \(1\text{/}(x-2)\) has a limit of \(\text{+\infty :}\)
\[\underset{x\to {2}^{-}}{\text{lim}}\frac{x-3}{{x}^{2}-2x}=\text{+}\infty .\]
Evaluate \(\underset{x\to 1}{\text{lim}}\frac{x+2}{{(x-1)}^{2}}.\)

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+∞
Apply the squeeze theorem to evaluate \(\underset{x\to 0}{\text{lim}}x\ \text{cos}\ x.\)

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Because \(-1\le \text{cos}\ x\le 1\) for all x, we have \(-|x|\le x\ \text{cos}\ x\le |x|\). Since \(\underset{x\to 0}{\text{lim}}(-|x|)=0=\underset{x\to 0}{\text{lim}}|x|,\) from the squeeze theorem, we obtain \(\underset{x\to 0}{\text{lim}}x\ \text{cos}\ x=0.\) The graphs of \(f(x)=-|x|,g(x)=x\ \text{cos}\ x,\) and \(h(x)=|x|\) are shown in .
Use the squeeze theorem to evaluate \(\underset{x\to 0}{\text{lim}}{x}^{2}\text{sin}\frac{1}{x}.\)

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0
Evaluate \(\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\theta }.\)

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In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:
\[\begin{array}{ll}\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\theta } & =\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\theta }\cdot \frac{1+\text{cos}\ \theta }{1+\text{cos}\ \theta } \\ & =\underset{\theta \to 0}{\text{lim}}\frac{1-{\text{cos}}^{2}\theta }{\theta (1+\text{cos}\ \theta )} \\ & =\underset{\theta \to 0}{\text{lim}}\frac{{\text{sin}}^{2}\theta }{\theta (1+\text{cos}\ \theta )} \\ & =\underset{\theta \to 0}{\text{lim}}\frac{\text{sin}\ \theta }{\theta }\cdot \frac{\text{sin}\ \theta }{1+\text{cos}\ \theta } \\ & =1\cdot \frac{0}{2}=0.\end{array}\]
Therefore,
\[\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\theta }=0.\]
Evaluate \(\underset{\theta \to 0}{\text{lim}}\frac{1-\text{cos}\ \theta }{\text{sin}\ \theta }.\)

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0
\(\underset{x\to 0}{\text{lim}}(4{x}^{2}-2x+3)\)

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Use constant multiple law and difference law: \(\underset{x\to 0}{\text{lim}}(4{x}^{2}-2x+3)=4\underset{x\to 0}{\text{lim}}{x}^{2}-2\underset{x\to 0}{\text{lim}}x+\underset{x\to 0}{\text{lim}}3=3\)
\(\underset{x\to 1}{\text{lim}}\frac{{x}^{3}+3{x}^{2}+5}{4-7x}\)
\(\underset{x\to -2}{\text{lim}}\sqrt{{x}^{2}-6x+3}\)

Otkrij odgovor

Use root law: \(\underset{x\to -2}{\text{lim}}\sqrt{{x}^{2}-6x+3}=\sqrt{\underset{x\to -2}{\text{lim}}({x}^{2}-6x+3)}=\sqrt{19}\)
\(\underset{x\to -1}{\text{lim}}{(9x+1)}^{2}\)
\(\underset{x\to 7}{\text{lim}}{x}^{2}\)

Otkrij odgovor

49
\(\underset{x\to -2}{\text{lim}}(4{x}^{2}-1)\)
\(\underset{x\to 0}{\text{lim}}\frac{1}{1+\text{sin}\ x}\)

Otkrij odgovor

1
\(\underset{x\to 2}{\text{lim}}{e}^{2x-{x}^{2}}\)
\(\underset{x\to 1}{\text{lim}}\frac{2-7x}{x+6}\)

Otkrij odgovor

\(-\frac{5}{7}\)
\(\underset{x\to 3}{\text{lim}}\text{ln}{e}^{3x}\)
\(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}\)

Otkrij odgovor

\(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}=\frac{16-16}{4-4}=\frac{0}{0};\) then, \(\underset{x\to 4}{\text{lim}}\frac{{x}^{2}-16}{x-4}=\underset{x\to 4}{\text{lim}}\frac{(x+4)(x-4)}{x-4}=8\)
\(\underset{x\to 2}{\text{lim}}\frac{x-2}{{x}^{2}-2x}\)
\(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}\)

Otkrij odgovor

\(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0};\) then, \(\underset{x\to 6}{\text{lim}}\frac{3x-18}{2x-12}=\underset{x\to 6}{\text{lim}}\frac{3(x-6)}{2(x-6)}=\frac{3}{2}\)
\(\underset{h\to 0}{\text{lim}}\frac{{(1+h)}^{2}-1}{h}\)
\(\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}\)

Otkrij odgovor

\(\underset{x\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}=\frac{9-9}{3-3}=\frac{0}{0};\) then, \(\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\text{lim}}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to 9}{\text{lim}}(\sqrt{t}+3)=6\)
\(\underset{h\to 0}{\text{lim}}\frac{\frac{1}{a+h}-\frac{1}{a}}{h},\) where a is a non-zero real-valued constant
\(\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\ \theta }{\text{tan}\ \theta }\)

Otkrij odgovor

\(\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\ \theta }{\text{tan}\ \theta }=\frac{\text{sin}\ \pi }{\text{tan}\ \pi }=\frac{0}{0};\) then, \(\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\ \theta }{\text{tan}\ \theta }=\underset{\theta \to \pi }{\text{lim}}\frac{\text{sin}\ \theta }{\frac{\text{sin}\ \theta }{\text{cos}\ \theta }}=\underset{\theta \to \pi }{\text{lim}}\text{cos}\ \theta =-1\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The Limit Laws

Recognize the basic limit laws.
Use the limit laws to evaluate the limit of a function.
Evaluate the limit of a function by factoring.
Use the limit laws to evaluate the limit of a polynomial or rational function.
Evaluate the limit of a function by factoring or by using conjugates.
Evaluate the limit of a function by using the squeeze theorem.
The limit of
The limit of a constant is that constant:

Questions people ask

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.

Pokušaj sam.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Više u Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests