maths.free › Calculus › 4. Applications of Derivatives › Limits at Infinity and Asymptotes
Limits at Infinity and Asymptotes
Calculate the limit of a function as
Limits at Infinity
We begin by examining what it means for a function to have a finite limit at infinity. Then we study the idea of a function with an infinite limit at infinity. Back in Introduction to Functions and Graphs, we looked at vertical asymptotes; in this section we deal with horizontal and oblique asymptotes.
Sometimes the values of a function \(f\) become arbitrarily large as \(x\to \infty\) (or as \(x\to \text{-}\infty ).\) In this case, we write \(\underset{x\to \infty }{\text{lim}}f(x)=\infty\) (or \(\underset{x\to \text{-}\infty }{\text{lim}}f(x)=\infty ).\) On the other hand, if the values of \(f\) are negative but become arbitrarily large in magnitude as \(x\to \infty\) (or as \(x\to \text{-}\infty ),\) we write \(\underset{x\to \infty }{\text{lim}}f(x)=\text{-}\infty\) (or \(\underset{x\to \text{-}\infty }{\text{lim}}f(x)=\text{-}\infty ).\)
For example, consider the function \(f(x)={x}^{3}.\) As seen in and , as \(x\to \infty\) the values \(f(x)\) become arbitrarily large. Therefore, \(\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .\) On the other hand, as \(x\to \text{-}\infty ,\) the values of \(f(x)={x}^{3}\) are negative but become arbitrarily large in magnitude. Consequently, \(\underset{x\to \text{-}\infty }{\text{lim}}{x}^{3}=\text{-}\infty .\)
| \(x\) | \(10\) | \(20\) | \(50\) | \(100\) | \(1000\) |
| \({x}^{3}\) | \(1000\) | \(8000\) | \(125,000\) | \(1,000,000\) | \(1,000,000,000\) |
| \(x\) | \(-10\) | \(-20\) | \(-50\) | \(-100\) | \(-1000\) |
| \({x}^{3}\) | \(-1000\) | \(-8000\) | \(-125,000\) | \(-1,000,000\) | \(-1,000,000,000\) |
Condensed — the full section is in OpenStax Calculus Volume 1.
End Behavior
The behavior of a function as \(x\to \text{\pm }\infty\) is called the function’s end behavior. At each of the function’s ends, the function could exhibit one of the following types of behavior:
- The function \(f(x)\) approaches a horizontal asymptote \(y=L.\)
- The function \(f(x)\to \infty\) or \(f(x)\to \text{-}\infty .\)
- The function does not approach a finite limit, nor does it approach \(\infty\) or \(\text{-}\infty .\) In this case, the function may have some oscillatory behavior.
Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.
Condensed — the full section is in OpenStax Calculus Volume 1.
Guidelines for Drawing the Graph of a Function
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Condensed — the full section is in OpenStax Calculus Volume 1.
Key Concepts
- The limit of \(f(x)\) is \(L\) as \(x\to \infty\) (or as \(x\to \text{-}\infty )\) if the values \(f(x)\) become arbitrarily close to \(L\) as \(x\) becomes sufficiently large.
- The limit of \(f(x)\) is \(\infty\) as \(x\to \infty\) if \(f(x)\) becomes arbitrarily large as \(x\) becomes sufficiently large. The limit of \(f(x)\) is \(\text{-}\infty\) as \(x\to \infty\) if \(f(x)<0\) and \(|f(x)|\) becomes arbitrarily large as \(x\) becomes sufficiently large. We can define the limit of \(f(x)\) as \(x\) approaches \(\text{-}\infty\) similarly.
- For a polynomial function \(p(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{\ldots }+{a}_{1}x+{a}_{0},\) where \({a}_{n}\ne 0,\) the end behavior is determined by the leading term \({a}_{n}{x}^{n}.\) If \(n\ne 0,\) \(p(x)\) approaches \(\infty\) or \(\text{-}\infty\) at each end.
- For a rational function \(f(x)=\frac{p(x)}{q(x)},\) the end behavior is determined by the relationship between the degree of \(p\) and the degree of \(q.\) If the degree of \(p\) is less than the degree of \(q,\) the line \(y=0\) is a horizontal asymptote for \(f.\) If the degree of \(p\) is equal to the degree of \(q,\) then the line \(y=\frac{{a}_{n}}{{b}_{n}}\) is a horizontal asymptote, where \({a}_{n}\) and \({b}_{n}\) are the leading coefficients of \(p\) and \(q,\) respectively. If the degree of \(p\) is greater than the degree of \(q,\) then \(f\) approaches \(\infty\) or \(\text{-}\infty\) at each end.
Limits at Infinity and Asymptotes
For the following exercises, examine the graphs. Identify where the vertical asymptotes are located.
For the following functions \(f(x),\) determine whether there is an asymptote at \(x=a.\) Justify your answer without graphing on a calculator.
For the following exercises, evaluate the limit.
For the following exercises, find the horizontal and vertical asymptotes.
For the following exercises, construct a function \(f(x)\) that has the given asymptotes.
For the following exercises, graph the function on a graphing calculator on the window \(x=[-5,5]\) and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.
For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
For each of the following functions \(f,\) evaluate \(\underset{x\to \infty }{\text{lim}}f(x)\) and \(\underset{x\to \text{-}\infty }{\text{lim}}f(x).\) Determine the horizontal asymptote(s) for \(f.\)
- \(f(x)=5-\frac{2}{{x}^{2}}\)
- \(f(x)=\frac{\text{sin}\ x}{x}\)
- \(f(x)={\text{tan}}^{-1}(x)\)
Revelar a resposta
- Using the algebraic limit laws, we have \(\underset{x\to \infty }{\text{lim}}(5-\frac{2}{{x}^{2}})=\underset{x\to \infty }{\text{lim}}5-2(\underset{x\to \infty }{\text{lim}}\frac{1}{x}).(\underset{x\to \infty }{\text{lim}}\frac{1}{x})=5-2\cdot 0=5.\)
Similarly, \(\underset{x\to -\infty }{\text{lim}}f(x)=5.\) Therefore, \(f(x)=5-\frac{2}{{x}^{2}}\) has a horizontal asymptote of \(y=5\) and \(f\) approaches this horizontal asymptote as \(x\to \text{\pm }\infty\) as shown in the following graph.
- Since \(-1\le \text{sin}\ x\le 1\) for all \(x,\) we have
\[\frac{-1}{x}\le \frac{\text{sin}\ x}{x}\le \frac{1}{x}\]
for all \(x\ne 0.\) Also, since
\[\underset{x\to \infty }{\text{lim}}\frac{-1}{x}=0=\underset{x\to \infty }{\text{lim}}\frac{1}{x},\]
we can apply the squeeze theorem to conclude that
\[\underset{x\to \infty }{\text{lim}}\frac{\text{sin}\ x}{x}=0.\]
Similarly,
\[\underset{x\to \text{-}\infty }{\text{lim}}\frac{\text{sin}\ x}{x}=0.\]
Thus, \(f(x)=\frac{\text{sin}\ x}{x}\) has a horizontal asymptote of \(y=0\) and \(f(x)\) approaches this horizontal asymptote as \(x\to \text{\pm }\infty\) as shown in the following graph.
- To evaluate \(\underset{x\to \infty }{\text{lim}}{\text{tan}}^{-1}(x)\) and \(\underset{x\to \text{-}\infty }{\text{lim}}{\text{tan}}^{-1}(x),\) we first consider the graph of \(y=\text{tan}(x)\) over the interval \((\text{-}\pi \text{/}2,\pi \text{/}2)\) as shown in the following graph.
Since
\[\underset{x\to {(\pi \text{/}2)}^{-}}{\text{lim}}\text{tan}\ x=\infty ,\]it follows that
\[\underset{x\to \infty }{\text{lim}}{\text{tan}}^{-1}(x)=\frac{\pi }{2}.\]Similarly, since
\[\underset{x\to {(-\pi \text{/}2)}^{+}}{\text{lim}}\text{tan}\ x=\text{-}\infty ,\]it follows that
\[\underset{x\to \text{-}\infty }{\text{lim}}{\text{tan}}^{-1}(x)=-\frac{\pi }{2}.\]As a result, \(y=\frac{\pi }{2}\) and \(y=-\frac{\pi }{2}\) are horizontal asymptotes of \(f(x)={\text{tan}}^{-1}(x)\) as shown in the following graph.
-
Evaluate \(\underset{x\to \text{-}\infty }{\text{lim}}(3+\frac{4}{x})\) and \(\underset{x\to \infty }{\text{lim}}(3+\frac{4}{x}).\) Determine the horizontal asymptotes of \(f(x)=3+\frac{4}{x},\) if any.
Revelar a resposta
Both limits are \(3.\) The line \(y=3\) is a horizontal asymptote.
-
Use the formal definition of limit at infinity to prove that \(\underset{x\to \infty }{\text{lim}}(2+\frac{1}{x})=2.\)
Revelar a resposta
Let \(\epsilon >0.\) Let \(N=\frac{1}{\epsilon }.\) Therefore, for all \(x>N,\) we have
\[|2+\frac{1}{x}-2|=|\frac{1}{x}|=\frac{1}{x}<\frac{1}{N}=\epsilon \text{.}\] -
Use the formal definition of limit at infinity to prove that \(\underset{x\to \infty }{\text{lim}}(3-\frac{1}{{x}^{2}})=3.\)
Revelar a resposta
Let \(\epsilon >0.\) Let \(N=\frac{1}{\sqrt{\epsilon }}.\) Therefore, for all \(x>N,\) we have
\(|3-\frac{1}{{x}^{2}}-3|=\frac{1}{{x}^{2}}<\frac{1}{{N}^{2}}=\epsilon\)
Therefore, \(\underset{x\to \infty }{\text{lim}}(3-1\text{/}{x}^{2})=3.\)
-
Use the formal definition of infinite limit at infinity to prove that \(\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .\)
Revelar a resposta
Let \(M>0.\) Let \(N=\sqrt[3]{M}.\) Then, for all \(x>N,\) we have
\[{x}^{3}>{N}^{3}={(\sqrt[3]{M})}^{3}=M.\]Therefore, \(\underset{x\to \infty }{\text{lim}}{x}^{3}=\infty .\)
-
Use the formal definition of infinite limit at infinity to prove that \(\underset{x\to \infty }{\text{lim}}3{x}^{2}=\infty .\)
Revelar a resposta
Let \(M>0.\) Let \(N=\sqrt{\frac{M}{3}}.\) Then, for all \(x>N,\) we have
\(3{x}^{2}>3{N}^{2}=3{(\sqrt{\frac{M}{3}})}^{2}{}^{2}=\frac{3M}{3}=M\)
-
For each function \(f,\) evaluate \(\underset{x\to \infty }{\text{lim}}f(x)\) and \(\underset{x\to \text{-}\infty }{\text{lim}}f(x).\)
- \(f(x)=-5{x}^{3}\)
- \(f(x)=2{x}^{4}\)
Revelar a resposta
- Since the coefficient of \({x}^{3}\) is \(-5,\) the graph of \(f(x)=-5{x}^{3}\) involves a vertical stretch and reflection of the graph of \(y={x}^{3}\) about the \(x\)-axis. Therefore, \(\underset{x\to \infty }{\text{lim}}(-5{x}^{3})=\text{-}\infty\) and \(\underset{x\to \text{-}\infty }{\text{lim}}(-5{x}^{3})=\infty .\)
- Since the coefficient of \({x}^{4}\) is \(2,\) the graph of \(f(x)=2{x}^{4}\) is a vertical stretch of the graph of \(y={x}^{4}.\) Therefore, \(\underset{x\to \infty }{\text{lim}}2{x}^{4}=\infty\) and \(\underset{x\to \text{-}\infty }{\text{lim}}2{x}^{4}=\infty .\)
-
Let \(f(x)=-3{x}^{4}.\) Find \(\underset{x\to \infty }{\text{lim}}f(x).\)
Revelar a resposta
\(\text{-}\infty\)
-
For each of the following functions, determine the limits as \(x\to \infty\) and \(x\to \text{-}\infty .\) Then, use this information to describe the end behavior of the function.
- \(f(x)=\frac{3x-1}{2x+5}\) (Note: The degree of the numerator and the denominator are the same.)
- \(f(x)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}\) (Note: The degree of numerator is less than the degree of the denominator.)
- \(f(x)=\frac{3{x}^{2}+4x}{x+2}\) (Note: The degree of numerator is greater than the degree of the denominator.)
Revelar a resposta
- The highest power of \(x\) in the denominator is \(x.\) Therefore, dividing the numerator and denominator by \(x\) and applying the algebraic limit laws, we see that
\[\begin{array}{ll}\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3x-1}{2x+5} & =\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3-1\text{/}x}{2+5\text{/}x} \\ & =\frac{\underset{x\to \text{\pm }\infty }{\text{lim}}(3-1\text{/}x)}{\underset{x\to \text{\pm }\infty }{\text{lim}}(2+5\text{/}x)} \\ & =\frac{\underset{x\to \text{\pm }\infty }{\text{lim}}3-\underset{x\to \text{\pm }\infty }{\text{lim}}1\text{/}x}{\underset{x\to \text{\pm }\infty }{\text{lim}}2+\underset{x\to \text{\pm }\infty }{\text{lim}}5\text{/}x} \\ & =\frac{3-0}{2+0}=\frac{3}{2}.\end{array}\]
Since \(\underset{x\to \text{\pm }\infty }{\text{lim}}f(x)=\frac{3}{2},\) we know that \(y=\frac{3}{2}\) is a horizontal asymptote for this function as shown in the following graph.
- Since the largest power of \(x\) appearing in the denominator is \({x}^{3},\) divide the numerator and denominator by \({x}^{3}.\) After doing so and applying algebraic limit laws, we obtain
\[\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}=\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3\text{/}x+2\text{/}{x}^{2}}{4-5\text{/}{x}^{2}+7\text{/}{x}^{3}}=\frac{3(0)+2(0)}{4-5(0)+7(0)}=0.\]
Therefore \(f\) has a horizontal asymptote of \(y=0\) as shown in the following graph.
- Dividing the numerator and denominator by \(x,\) we have
\[\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3{x}^{2}+4x}{x+2}=\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3x+4}{1+2\text{/}x}.\]
As \(x\to \text{\pm }\infty ,\) the denominator approaches \(1.\) As \(x\to \infty ,\) the numerator approaches \(+\infty .\) As \(x\to \text{-}\infty ,\) the numerator approaches \(\text{-}\infty .\) Therefore \(\underset{x\to \infty }{\text{lim}}f(x)=\infty ,\) whereas \(\underset{x\to \text{-}\infty }{\text{lim}}f(x)=\text{-}\infty\) as shown in the following figure.
-
Evaluate \(\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{3{x}^{2}+2x-1}{5{x}^{2}-4x+7}\) and use these limits to determine the end behavior of \(f(x)=\frac{3{x}^{2}+2x-1}{5{x}^{2}-4x+7}.\)
Revelar a resposta
\(\frac{3}{5}\)
-
Find the limits as \(x\to \infty\) and \(x\to \text{-}\infty\) for \(f(x)=\frac{3x-2}{\sqrt{4{x}^{2}+5}}\) and describe the end behavior of \(f.\)
Revelar a resposta
Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of \(x.\) To determine the appropriate power of \(x,\) consider the expression \(\sqrt{4{x}^{2}+5}\) in the denominator. Since
\[\sqrt{4{x}^{2}+5}\approx \sqrt{4{x}^{2}}=2|x|\]for large values of \(x\) in effect \(x\) appears just to the first power in the denominator. Therefore, we divide the numerator and denominator by \(|x|.\) Then, using the fact that \(|x|=x\) for \(x>0,\) \(|x|=\text{-}x\) for \(x<0,\) and \(|x|=\sqrt{{x}^{2}}\) for all \(x,\) we calculate the limits as follows:
\[\begin{array}{lll}\underset{x\to \infty }{\text{lim}}\frac{3x-2}{\sqrt{4{x}^{2}+5}} & = & \underset{x\to \infty }{\text{lim}}\frac{(1\text{/}|x|)(3x-2)}{(1\text{/}|x|)\sqrt{4{x}^{2}+5}} \\ & = & \underset{x\to \infty }{\text{lim}}\frac{(1\text{/}x)(3x-2)}{\sqrt{(1\text{/}{x}^{2})(4{x}^{2}+5)}} \\ & = & \underset{x\to \infty }{\text{lim}}\frac{3-2\text{/}x}{\sqrt{4+5\text{/}{x}^{2}}}=\frac{3}{\sqrt{4}}=\frac{3}{2} \\ \underset{x\to \text{-}\infty }{\text{lim}}\frac{3x-2}{\sqrt{4{x}^{2}+5}} & = & \underset{x\to \text{-}\infty }{\text{lim}}\frac{(1\text{/}|x|)(3x-2)}{(1\text{/}|x|)\sqrt{4{x}^{2}+5}} \\ & = & \underset{x\to \text{-}\infty }{\text{lim}}\frac{(-1\text{/}x)(3x-2)}{\sqrt{(1\text{/}{x}^{2})(4{x}^{2}+5)}} \\ & = & \underset{x\to \text{-}\infty }{\text{lim}}\frac{-3+2\text{/}x}{\sqrt{4+5\text{/}{x}^{2}}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.\end{array}\]Therefore, \(f(x)\) approaches the horizontal asymptote \(y=\frac{3}{2}\) as \(x\to \infty\) and the horizontal asymptote \(y=-\frac{3}{2}\) as \(x\to \text{-}\infty\) as shown in the following graph.
-
Evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\sqrt{3{x}^{2}+4}}{x+6}.\)
Revelar a resposta
\(\text{\pm }\sqrt{3}\)
-
Find the limits as \(x\to \infty\) and \(x\to \text{-}\infty\) for \(f(x)=\frac{(2+3{e}^{x})}{(7-5{e}^{x})}\) and describe the end behavior of \(f.\)
Revelar a resposta
To find the limit as \(x\to \infty ,\) divide the numerator and denominator by \({e}^{x}\text{:}\)
\[\begin{array}{ll}\underset{x\to \infty }{\text{lim}}f(x) & =\underset{x\to \infty }{\text{lim}}\frac{2+3{e}^{x}}{7-5{e}^{x}} \\ & =\underset{x\to \infty }{\text{lim}}\frac{(2\text{/}{e}^{x})+3}{(7\text{/}{e}^{x})-5}.\end{array}\]As shown in , \({e}^{x}\to \infty\) as \(x\to \infty .\) Therefore,
\[\underset{x\to \infty }{\text{lim}}\frac{2}{{e}^{x}}=0=\underset{x\to \infty }{\text{lim}}\frac{7}{{e}^{x}}.\]We conclude that \(\underset{x\to \infty }{\text{lim}}f(x)=-\frac{3}{5},\) and the graph of \(f\) approaches the horizontal asymptote \(y=-\frac{3}{5}\) as \(x\to \infty .\) To find the limit as \(x\to \text{-}\infty ,\) use the fact that \({e}^{x}\to 0\) as \(x\to \text{-}\infty\) to conclude that \(\underset{x\to \infty }{\text{lim}}f(x)=\frac{2}{7},\) and therefore the graph of approaches the horizontal asymptote \(y=\frac{2}{7}\) as \(x\to \text{-}\infty .\)
-
Find the limits as \(x\to \infty\) and \(x\to \text{-}\infty\) for \(f(x)=\frac{(3{e}^{x}-4)}{(5{e}^{x}+2)}.\)
Revelar a resposta
\(\underset{x\to \infty }{\text{lim}}f(x)=\frac{3}{5},\) \(\underset{x\to \text{-}\infty }{\text{lim}}f(x)=-2\)
-
Sketch a graph of \(f(x)={(x-1)}^{2}(x+2).\)
Revelar a resposta
Step 1. Since \(f\) is a polynomial, the domain is the set of all real numbers.
Step 2. When \(x=0,f(x)=2.\) Therefore, the \(y\)-intercept is \((0,2).\) To find the \(x\)-intercepts, we need to solve the equation \({(x-1)}^{2}(x+2)=0,\) gives us the \(x\)-intercepts \((1,0)\) and \((-2,0)\)
Step 3. We need to evaluate the end behavior of \(f.\) As \(x\to \infty ,\) \({(x-1)}^{2}\to \infty\) and \((x+2)\to \infty .\) Therefore, \(\underset{x\to \infty }{\text{lim}}f(x)=\infty .\) As \(x\to \text{-}\infty ,\) \({(x-1)}^{2}\to \infty\) and \((x+2)\to \text{-}\infty .\) Therefore, \(\underset{x\to \text{-}\infty }{\text{lim}}f(x)=\text{-}\infty .\) To get even more information about the end behavior of \(f,\) we can multiply the factors of \(f.\) When doing so, we see that
\[f(x)={(x-1)}^{2}(x+2)={x}^{3}-3x+2.\]Since the leading term of \(f\) is \({x}^{3},\) we conclude that \(f\) behaves like \(y={x}^{3}\) as \(x\to \text{\pm }\infty .\)
Step 4. Since \(f\) is a polynomial function, it does not have any vertical asymptotes.
Step 5. The first derivative of \(f\) is
\[{f}^{'}(x)=3{x}^{2}-3.\]Therefore, \(f\) has two critical points: \(x=1,-1.\) Divide the interval \((\text{-}\infty ,\infty )\) into the three smaller intervals: \((\text{-}\infty ,-1),\) \((-1,1),\) and \((1,\infty ).\) Then, choose test points \(x=-2,\) \(x=0,\) and \(x=2\) from these intervals and evaluate the sign of \({f}^{'}(x)\) at each of these test points, as shown in the following table.
Interval Test Point Sign of Derivative \(f'(x)=3{x}^{2}-3=3(x-1)(x+1)\) Conclusion \((\text{-}\infty ,-1)\) \(x=-2\) \((\text{+})(\text{-})(\text{-})=+\) \(f\) is increasing. \((-1,1)\) \(x=0\) \((\text{+})(\text{-})(\text{+})=\text{-}\) \(f\) is decreasing. \((1,\infty )\) \(x=2\) \((\text{+})(\text{+})(\text{+})=+\) \(f\) is increasing. From the table, we see that \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1.\) Evaluating \(f(x)\) at those two points, we find that the local maximum value is \(f(-1)=4\) and the local minimum value is \(f(1)=0.\)
Step 6. The second derivative of \(f\) is
\[{f}^{″}(x)=6x.\]The second derivative is zero at \(x=0.\) Therefore, to determine the concavity of \(f,\) divide the interval \((\text{-}\infty ,\infty )\) into the smaller intervals \((\text{-}\infty ,0)\) and \((0,\infty ),\) and choose test points \(x=-1\) and \(x=1\) to determine the concavity of \(f\) on each of these smaller intervals as shown in the following table.
Interval Test Point Sign of \({f}^{″}(x)=6x\) Conclusion \((\text{-}\infty ,0)\) \(x=-1\) \(-\) \(f\) is concave down. \((0,\infty )\) \(x=1\) \(+\) \(f\) is concave up. We note that the information in the preceding table confirms the fact, found in step \(5,\) that \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1.\) In addition, the information found in step \(5\)—namely, \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=1,\) and \({f}^{'}(x)=0\) at those points—combined with the fact that \({f}^{″}\) changes sign only at \(x=0\) confirms the results found in step \(6\) on the concavity of \(f.\)
Combining this information, we arrive at the graph of \(f(x)={(x-1)}^{2}(x+2)\) shown in the following graph.
-
Sketch a graph of \(f(x)={(x-1)}^{3}(x+2).\)
Revelar a resposta
-
Sketch the graph of \(f(x)=\frac{{x}^{2}}{(1-{x}^{2})}\text{.}\)
Revelar a resposta
Step 1. The function \(f\) is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers \(x\) except \(x=\text{\pm }1.\)
Step 2. Find the intercepts. If \(x=0,\) then \(f(x)=0,\) so \(0\) is an intercept. If \(y=0,\) then \(\frac{{x}^{2}}{(1-{x}^{2})}=0,\) which implies \(x=0.\) Therefore, \((0,0)\) is the only intercept.
Step 3. Evaluate the limits at infinity. Since \(f\) is a rational function, divide the numerator and denominator by the highest power in the denominator: \({x}^{2}.\) We obtain
\[\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\underset{x\to \text{\pm }\infty }{\text{lim}}\frac{1}{\frac{1}{{x}^{2}}-1}=-1.\]Therefore, \(f\) has a horizontal asymptote of \(y=-1\) as \(x\to \infty\) and \(x\to \text{-}\infty .\)
Step 4. To determine whether \(f\) has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when \(x=\text{\pm }1.\) To determine whether the lines \(x=1\) or \(x=-1\) are vertical asymptotes of \(f,\) evaluate \(\underset{x\to 1}{\text{lim}}f(x)\) and \(\underset{x\to \text{-}1}{\text{lim}}f(x).\) By looking at each one-sided limit as \(x\to 1,\) we see that
\[\underset{x\to {1}^{+}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\text{-}\infty \ \text{and}\ \underset{x\to {1}^{-}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\infty .\]In addition, by looking at each one-sided limit as \(x\to \text{-}1,\) we find that
\[\underset{x\to \text{-}{1}^{+}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\infty \ \text{and}\ \underset{x\to \text{-}{1}^{-}}{\text{lim}}\frac{{x}^{2}}{1-{x}^{2}}=\text{-}\infty .\]Step 5. Calculate the first derivative:
\[{f}^{'}(x)=\frac{(1-{x}^{2})(2x)-{x}^{2}(-2x)}{{(1-{x}^{2})}^{2}}=\frac{2x}{{(1-{x}^{2})}^{2}}.\]Critical points occur at points \(x\) where \({f}^{'}(x)=0\) or \({f}^{'}(x)\) is undefined. We see that \({f}^{'}(x)=0\) when \(x=0.\) The derivative \({f}^{'}\) is not undefined at any point in the domain of \(f.\) However, \(x=\text{\pm }1\) are not in the domain of \(f.\) Therefore, to determine where \(f\) is increasing and where \(f\) is decreasing, divide the interval \((\text{-}\infty ,\infty )\) into four smaller intervals: \((\text{-}\infty ,-1),\) \((-1,0),\) \((0,1),\) and \((1,\infty ),\) and choose a test point in each interval to determine the sign of \({f}^{'}(x)\) in each of these intervals. The values \(x=-2,\) \(x=-\frac{1}{2},\) \(x=\frac{1}{2},\) and \(x=2\) are good choices for test points as shown in the following table.
Interval Test Point Sign of \({f}^{'}(x)=\frac{2x}{{(1-{x}^{2})}^{2}}\) Conclusion \((\text{-}\infty ,-1)\) \(x=-2\) \(\text{-}\text{/}+=\text{-}\) \(f\) is decreasing. \((-1,0)\) \(x=-1\text{/}2\) \(\text{-}\text{/}+=\text{-}\) \(f\) is decreasing. \((0,1)\) \(x=1\text{/}2\) \(+\text{/}+=+\) \(f\) is increasing. \((1,\infty )\) \(x=2\) \(+\text{/}+=+\) \(f\) is increasing. From this analysis, we conclude that \(f\) has a local minimum at \(x=0\) but no local maximum.
Step 6. Calculate the second derivative:
\[\begin{array}{ll}{f}^{″}(x) & =\frac{{(1-{x}^{2})}^{2}(2)-2x(2(1-{x}^{2})(-2x))}{{(1-{x}^{2})}^{4}} \\ & =\frac{(1-{x}^{2})[2(1-{x}^{2})+8{x}^{2}]}{{(1-{x}^{2})}^{4}} \\ & =\frac{2(1-{x}^{2})+8{x}^{2}}{{(1-{x}^{2})}^{3}} \\ & =\frac{6{x}^{2}+2}{{(1-{x}^{2})}^{3}}.\end{array}\]To determine the intervals where \(f\) is concave up and where \(f\) is concave down, we first need to find all points \(x\) where \({f}^{″}(x)=0\) or \({f}^{″}(x)\) is undefined. Since the numerator \(6{x}^{2}+2\ne 0\) for any \(x,\) \({f}^{″}(x)\) is never zero. Furthermore, \({f}^{″}\) is not undefined for any \(x\) in the domain of \(f.\) However, as discussed earlier, \(x=\text{\pm }1\) are not in the domain of \(f.\) Therefore, to determine the concavity of \(f,\) we divide the interval \((\text{-}\infty ,\infty )\) into the three smaller intervals \((\text{-}\infty ,-1),\) \((-1,-1),\) and \((1,\infty ),\) and choose a test point in each of these intervals to evaluate the sign of \({f}^{″}(x).\) in each of these intervals. The values \(x=-2,\) \(x=0,\) and \(x=2\) are possible test points as shown in the following table.
Interval Test Point Sign of \({f}^{″}(x)=\frac{6{x}^{2}+2}{{(1-{x}^{2})}^{3}}\) Conclusion \((\text{-}\infty ,-1)\) \(x=-2\) \(+\text{/}-=\text{-}\) \(f\) is concave down. \((-1,-1)\) \(x=0\) \(+\text{/}+=+\) \(f\) is concave up. \((1,\infty )\) \(x=2\) \(+\text{/}-=\text{-}\) \(f\) is concave down. Combining all this information, we arrive at the graph of \(f\) shown below. Note that, although \(f\) changes concavity at \(x=-1\) and \(x=1,\) there are no inflection points at either of these places because \(f\) is not continuous at \(x=-1\) or \(x=1.\)
-
Sketch a graph of \(f(x)=\frac{(3x+5)}{(8+4x)}.\)
Revelar a resposta
-
Sketch the graph of \(f(x)=\frac{{x}^{2}}{(x-1)}\)
Revelar a resposta
Step 1. The domain of \(f\) is the set of all real numbers \(x\) except \(x=1.\)
Step 2. Find the intercepts. We can see that when \(x=0,\) \(f(x)=0,\) so \((0,0)\) is the only intercept.
Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, \(f\) must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write
\[f(x)=\frac{{x}^{2}}{x-1}=x+1+\frac{1}{x-1}.\]Since \(1\text{/}(x-1)\to 0\) as \(x\to \text{\pm }\infty ,\) \(f(x)\) approaches the line \(y=x+1\) as \(x\to \text{\pm }\infty .\) The line \(y=x+1\) is an oblique asymptote for \(f.\)
Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at \(x=1.\) Looking at both one-sided limits as \(x\to 1,\) we find
\[\underset{x\to {1}^{+}}{\text{lim}}\frac{{x}^{2}}{x-1}=\infty \ \text{and}\ \underset{x\to {1}^{-}}{\text{lim}}\frac{{x}^{2}}{x-1}=\text{-}\infty .\]Therefore, \(x=1\) is a vertical asymptote, and we have determined the behavior of \(f\) as \(x\) approaches \(1\) from the right and the left.
Step 5. Calculate the first derivative:
\[{f}^{'}(x)=\frac{(x-1)(2x)-{x}^{2}(1)}{{(x-1)}^{2}}=\frac{{x}^{2}-2x}{{(x-1)}^{2}}.\]We have \({f}^{'}(x)=0\) when \({x}^{2}-2x=x(x-2)=0.\) Therefore, \(x=0\) and \(x=2\) are critical points. Since \(f\) is undefined at \(x=1,\) we need to divide the interval \((\text{-}\infty ,\infty )\) into the smaller intervals \((\text{-}\infty ,0),\) \((0,1),\) \((1,2),\) and \((2,\infty ),\) and choose a test point from each interval to evaluate the sign of \({f}^{'}(x)\) in each of these smaller intervals. For example, let \(x=-1,\) \(x=\frac{1}{2},\) \(x=\frac{3}{2},\) and \(x=3\) be the test points as shown in the following table.
Interval Test Point Sign of \(f'(x)=\frac{{x}^{2}-2x}{{(x-1)}^{2}}=\frac{x(x-2)}{{(x-1)}^{2}}\) Conclusion \((\text{-}\infty ,0)\) \(x=-1\) \((\text{-})(\text{-})\text{/}+=+\) \(f\) is increasing. \((0,1)\) \(x=1\text{/}2\) \((\text{+})(\text{-})\text{/}+=\text{-}\) \(f\) is decreasing. \((1,2)\) \(x=3\text{/}2\) \((\text{+})(\text{-})\text{/}+=\text{-}\) \(f\) is decreasing. \((2,\infty )\) \(x=3\) \((\text{+})(\text{+})\text{/}+=+\) \(f\) is increasing. From this table, we see that \(f\) has a local maximum at \(x=0\) and a local minimum at \(x=2.\) The value of \(f\) at the local maximum is \(f(0)=0\) and the value of \(f\) at the local minimum is \(f(2)=4.\) Therefore, \((0,0)\) and \((2,4)\) are important points on the graph.
Step 6. Calculate the second derivative:
\[\begin{array}{ll}{f}^{″}(x) & =\frac{{(x-1)}^{2}(2x-2)-({x}^{2}-2x)(2(x-1))}{{(x-1)}^{4}} \\ & =\frac{(x-1)[(x-1)(2x-2)-2({x}^{2}-2x)]}{{(x-1)}^{4}} \\ & =\frac{(x-1)(2x-2)-2({x}^{2}-2x)}{{(x-1)}^{3}} \\ & =\frac{2{x}^{2}-4x+2-(2{x}^{2}-4x)}{{(x-1)}^{3}} \\ & =\frac{2}{{(x-1)}^{3}}.\end{array}\]We see that \({f}^{″}(x)\) is never zero or undefined for \(x\) in the domain of \(f.\) Since \(f\) is undefined at \(x=1,\) to check concavity we just divide the interval \((\text{-}\infty ,\infty )\) into the two smaller intervals \((\text{-}\infty ,1)\) and \((1,\infty ),\) and choose a test point from each interval to evaluate the sign of \({f}^{″}(x)\) in each of these intervals. The values \(x=0\) and \(x=2\) are possible test points as shown in the following table.
Interval Test Point Sign of \({f}^{″}(x)=\frac{2}{{(x-1)}^{3}}\) Conclusion \((\text{-}\infty ,1)\) \(x=0\) \(+\text{/}-=\text{-}\) \(f\) is concave down. \((1,\infty )\) \(x=2\) \(+\text{/}+=+\) \(f\) is concave up. From the information gathered, we arrive at the following graph for \(f.\)
-
Find the oblique asymptote for \(f(x)=\frac{(3{x}^{3}-2x+1)}{(2{x}^{2}-4)}.\)
Revelar a resposta
\(y=\frac{3}{2}x\)
-
Sketch a graph of \(f(x)={(x-1)}^{2\text{/}3}.\)
Revelar a resposta
Step 1. Since the cube-root function is defined for all real numbers \(x\) and \({(x-1)}^{2\text{/}3}={(\sqrt[3]{x-1})}^{2},\) the domain of \(f\) is all real numbers.
Step 2: To find the \(y\)-intercept, evaluate \(f(0).\) Since \(f(0)=1,\) the \(y\)-intercept is \((0,1).\) To find the \(x\)-intercept, solve \({(x-1)}^{2\text{/}3}=0.\) The solution of this equation is \(x=1,\) so the \(x\)-intercept is \((1,0).\)
Step 3: Since \(\underset{x\to \text{\pm }\infty }{\text{lim}}{(x-1)}^{2\text{/}3}=\infty ,\) the function continues to grow without bound as \(x\to \infty\) and \(x\to \text{-}\infty .\)
Step 4: The function has no vertical asymptotes.
Step 5: To determine where \(f\) is increasing or decreasing, calculate \({f}^{'}.\) We find
\[{f}^{'}(x)=\frac{2}{3}{(x-1)}^{-1\text{/}3}=\frac{2}{3{(x-1)}^{1\text{/}3}}.\]This function is not zero anywhere, but it is undefined when \(x=1.\) Therefore, the only critical point is \(x=1.\) Divide the interval \((\text{-}\infty ,\infty )\) into the smaller intervals \((\text{-}\infty ,1)\) and \((1,\infty ),\) and choose test points in each of these intervals to determine the sign of \({f}^{'}(x)\) in each of these smaller intervals. Let \(x=0\) and \(x=2\) be the test points as shown in the following table.
Interval Test Point Sign of \({f}^{'}(x)=\frac{2}{3{(x-1)}^{1\text{/}3}}\) Conclusion \((\text{-}\infty ,1)\) \(x=0\) \(+\text{/}-=\text{-}\) \(f\) is decreasing. \((1,\infty )\) \(x=2\) \(+\text{/}+=+\) \(f\) is increasing. We conclude that \(f\) has a local minimum at \(x=1.\) Evaluating \(f\) at \(x=1,\) we find that the value of \(f\) at the local minimum is zero. Note that \({f}^{'}(1)\) is undefined, so to determine the behavior of the function at this critical point, we need to examine \(\underset{x\to 1}{\text{lim}}{f}^{'}(x).\) Looking at the one-sided limits, we have
\[\underset{x\to {1}^{+}}{\text{lim}}\frac{2}{3{(x-1)}^{1\text{/}3}}=\infty \ \text{and}\ \underset{x\to {1}^{-}}{\text{lim}}\frac{2}{3{(x-1)}^{1\text{/}3}}=\text{-}\infty .\]Therefore, \(f\) has a cusp at \(x=1.\)
Step 6: To determine concavity, we calculate the second derivative of \(f\text{:}\)
\[{f}^{″}(x)=-\frac{2}{9}{(x-1)}^{-4\text{/}3}=\frac{-2}{9{(x-1)}^{4\text{/}3}}.\]We find that \({f}^{″}(x)\) is defined for all \(x,\) but is undefined when \(x=1.\) Therefore, divide the interval \((\text{-}\infty ,\infty )\) into the smaller intervals \((\text{-}\infty ,1)\) and \((1,\infty ),\) and choose test points to evaluate the sign of \({f}^{″}(x)\) in each of these intervals. As we did earlier, let \(x=0\) and \(x=2\) be test points as shown in the following table.
Interval Test Point Sign of \({f}^{″}(x)=\frac{-2}{9{(x-1)}^{4\text{/}3}}\) Conclusion \((\text{-}\infty ,1)\) \(x=0\) \(\text{-}\text{/}+=\text{-}\) \(f\) is concave down. \((1,\infty )\) \(x=2\) \(\text{-}\text{/}+=\text{-}\) \(f\) is concave down. From this table, we conclude that \(f\) is concave down everywhere. Combining all of this information, we arrive at the following graph for \(f.\)
-
Consider the function \(f(x)=5-{x}^{2\text{/}3}.\) Determine the point on the graph where a cusp is located. Determine the end behavior of \(f.\)
Revelar a resposta
The function \(f\) has a cusp at \((0,5)\) \(\underset{x\to {0}^{-}}{\text{lim}}{f}^{'}(x)=\infty ,\) \(\underset{x\to {0}^{+}}{\text{lim}}{f}^{'}(x)=\text{-}\infty .\) For end behavior, \(\underset{x\to \text{\pm }\infty }{\text{lim}}f(x)=\text{-}\infty .\)
-
\(f(x)=\frac{x+1}{{x}^{2}+5x+4},a=-1\)
-
\(f(x)=\frac{x}{x-2},a=2\)
Revelar a resposta
Yes, there is a vertical asymptote
-
\(f(x)={(x+2)}^{3\text{/}2},a=-2\)
-
\(f(x)={(x-1)}^{-1\text{/}3},a=1\)
Revelar a resposta
Yes, there is vertical asymptote
-
\(f(x)=1+{x}^{-2\text{/}5},a=1\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{1}{3x+6}\)
Revelar a resposta
\(0\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{2x-5}{4x}\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}-2x+5}{x+2}\)
Revelar a resposta
\(\infty\)
-
\(\underset{x\to \text{-}\infty }{\text{lim}}\frac{3{x}^{3}-2x}{{x}^{2}+2x+8}\)
-
\(\underset{x\to \text{-}\infty }{\text{lim}}\frac{{x}^{4}-4{x}^{3}+1}{2-2{x}^{2}-7{x}^{4}}\)
Revelar a resposta
\(-\frac{1}{7}\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{3x}{\sqrt{{x}^{2}+1}}\)
-
\(\underset{x\to \text{-}\infty }{\text{lim}}\frac{\sqrt{4{x}^{2}-1}}{x+2}\)
Revelar a resposta
\(-2\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}\)
-
\(\underset{x\to \text{-}\infty }{\text{lim}}\frac{4x}{\sqrt{{x}^{2}-1}}\)
Revelar a resposta
\(-4\)
-
\(\underset{x\to \infty }{\text{lim}}\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)
-
\(f(x)=x-\frac{9}{x}\)
Revelar a resposta
Horizontal: none, vertical: \(x=0\)
-
\(f(x)=\frac{1}{1-{x}^{2}}\)
-
\(f(x)=\frac{{x}^{3}}{4-{x}^{2}}\)
Revelar a resposta
Horizontal: none, vertical: \(x=\text{\pm }2\)
Symbols used here
The non-negative number whose square (n-th power) is x.
Not a number: "grows without bound" in limits and intervals.
Both signs at once: x = 3 ± 2 means 5 and 1.
The two sides are different.
Least upper bound, greatest lower bound.
Ratio of a circle's circumference to its diameter, 3.14159…
2.71828…, the base whose exponential is its own derivative.
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
Add a_k for k = 1 up to n.
The value f(x) approaches as x approaches a.
Instantaneous rate of change; slope of the graph.
Antiderivative (indefinite) or signed area from a to b (definite).
Prime notation for derivatives with respect to x (or t).
Constants of integration fixed by initial conditions.
How to: Limits at Infinity and Asymptotes
- Calculate the limit of a function as
- Recognize a horizontal asymptote on the graph of a function.
- Estimate the end behavior of a function as
- Recognize an oblique asymptote on the graph of a function.
- Analyze a function and its derivatives to draw its graph.
- Using the algebraic limit laws, we have
- Since
- To evaluate
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.
Tente o seu próprio
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.
Mais em Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests