2.7 Limits at Infinity

So far we have considered limits as \(x\) approaches a number \(c\). It is also important to consider limits where \(x\) approaches \(\infty\) or \(-\infty\), which we refer to as limits at infinity. In applications, limits at infinity arise naturally when we describe the “long-term” behavior of a system.

The notation \(x\rightarrow\infty\) indicates that \(x\) increases without bound, and \(x\rightarrow -\infty\) indicates that \(x\) decreases (through negative values) without bound. We write

As before, “closer and closer” means that \(|f(x)-L|\) becomes arbitrarily small. In either case, the line \(y=L\) is called a horizontal asymptote. We use the notation \(x\rightarrow\pm\infty\) to indicate that we are considering both infinite limits, as \(x\rightarrow\infty\) and as \(x\rightarrow -\infty\).

Infinite limits describe the asymptotic behavior of a function, which is behavior of the graph as we move out to the right or the left.

Example 1

Discuss the asymptotic behavior in Figure 2.41.

Figure 2.41: The lines \(y=7\) and \(y = 3\) are horizontal asymptotes of \(g(x)\).

Solution The function \(g(x)\) approaches \(L=7\) as we move to the right and it approaches \(L=3\) as we move to left, so

\[\lim\limits_{x\rightarrow \infty} g(x) = 7 \text{ and } \lim\limits_{x\rightarrow -\infty} g(x) = 3\]

Accordingly, the lines \(y=7\) and \(y = 3\) are horizontal asymptotes of \(g(x)\).

A function may approach an infinite limit as \(x\rightarrow\pm\infty\). We write

\[\lim\limits_{x\rightarrow \infty} f(x) = \infty \text{ and } \lim\limits_{x\rightarrow -\infty} f(x) = \infty\]

if \(f(x)\) becomes arbitrarily large as \(x\rightarrow\pm\infty\) respectively. Similar notation is used if \(f(x)\) approaches \(-\infty\) as \(x\rightarrow\pm\infty\). For example, we see in Figure 2.42(A) that

\[\lim\limits_{x\rightarrow \infty} e^x = \infty \text{ and }\lim\limits_{x\rightarrow -\infty}e^x = 0 \]

Figure 2.42: The above graphs display different possibilities of behaviors of functions as \(x\rightarrow\pm\infty\).

101

However, limits at infinity do not always exist. For example, \(f(x)=\sin x\) oscillates indefinitely [Figure 2.42(B)], so

\[\lim\limits_{x\rightarrow \infty}\sin x \text{ and } \lim\limits_{x\rightarrow -\infty}\sin x\]

do not exist.

The limits at infinity of the power functions \(f(x)=x^n\) are easily determined. If \(n > 0\), then \(x^n\) certainly increases without bound as \(x\rightarrow\infty\), so (Figure 2.43)

\[\lim\limits_{x\rightarrow \infty}x^n = \infty \text{ and } \lim\limits_{x\rightarrow \infty} x^{-n} = \lim\limits_{x\rightarrow \infty} \frac{1}{x^n} = 0\]

Figure 2.43: The given graphs depict some of the possible long-term bahaviours of power functions.

To describe the limits as \(x\rightarrow-\infty\), assume that \(n\) is a whole number so that \(x^n\) is defined for \(x < 0\). If \(n\) is even, then \(x^n\) becomes large and positive as \(x\rightarrow-\infty\), and if \(n\) is odd, it becomes large and negative. In summary,

THEOREM 1

For all \(n > 0\),

\[\lim\limits_{x\rightarrow \infty}x^n = \infty \text{ and }\lim\limits_{x\rightarrow\pm\infty} x^{-n} = \lim\limits_{x\rightarrow \infty} \frac{1}{x^n} = 0\]

If \(n\) is a whole number,

\[\lim\limits_{x\rightarrow-\infty} x^n = \left\{\begin{aligned}\infty & \text{if }n\text{ is even}\\-\infty & \text{if }n\text{ is odd}\\ \end{aligned}\right.\text{ and } \lim\limits_{x\rightarrow\pm\infty} x^{-n} =0\]

The Basic Limit Laws (Theorem 1 in Section 2.7) are valid for limits at infinity. For example, the Sum and Constant Multiple Laws yield:

\[\begin{aligned}\lim\limits_{x\rightarrow \infty}(3-4x^{-3}+5x^{-5}) & = \lim\limits_{x\rightarrow \infty} 3 - 4\lim\limits_{x\rightarrow \infty}x^{-3} + 5\lim\limits_{x\rightarrow \infty}x^{-5}\\ & = 3 - 0 + 0 = 3\end{aligned}\]

Question 2.20 Infinite Limits Progress Check 1

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3
Correct.
Try again.
Incorrect.

102

Example 2

Calculate \(\lim\limits_{x\rightarrow\pm\infty} \dfrac{20x^2 - 3x}{3x^5 -4x^2 +5}\).

Solution It would be nice if we could apply the Quotient Law directly, but this law is valid only if the denominator has a finite, nonzero limit. Our limit has the indeterminate form \(\frac{\infty}{\infty}\) because

\[\lim\limits_{x\rightarrow\infty} 20x^2 - 3x = \infty \text{ and }\lim\limits_{x\rightarrow \infty} (3x^5 - 4x^2 + 5) = \infty\]

The way around this difficulty is to divide the numerator and denominator by \(x^5\) (the highest power of \(x\) in the denominator):

\[\dfrac{20x^2 - 3x}{3x^5 -4x^2 +5} = \dfrac{x^{-5}(20x^2 - 3x)}{x^{-5}(3x^5 -4x^2 +5)} = \dfrac{20x^{-3} - 3x^{-4}}{3 -4x^{-3} +5x^{-5}} \]

Now we can use the Quotient Law:

\[\lim\limits_{x\rightarrow\pm\infty} \dfrac{20x^2 - 3x}{3x^5 -4x^2 +5} = \dfrac{\lim\limits_{x\rightarrow\pm\infty}(20x^{-3} - 3x^{-4})}{\lim\limits_{x\rightarrow\pm\infty}(3 -4x^{-3} +5x^{-5})} = \frac{0}{3} = 0\]

In general, if

\[f(x) =\dfrac{a_nx^n + a_{n-1}x^{n-1}+\cdots+a_0}{b_mx^m + b_{m-1}x^{m-1}+\cdots+b_0}\]

where \(a_n,b_m\neq 0\), divide the numerator and denominator by \(x^m\):

\[\begin{aligned} f(x) & =\dfrac{a_nx^{n-m} + a_{n-1}x^{n-1-m}+\cdots+a_0x^{-m}}{b_m + b_{m-1}x^{-1}+\cdots+b_0x^{-m}}\\ & = x^{n-m}\left(\dfrac{a_n + a_{n-1}x^{-1}+\cdots+a_0x^{-n}}{b_m + b_{m-1}x^{-1}+\cdots+b_0x^{-m}}\right)\end{aligned}\]

The quotient in parenthesis approaches the finite limit \(\frac{a_n}{b_m}\) because

\[\begin{aligned}\lim\limits_{x\rightarrow\infty}\left(a_n + a_{n-1}x^{-1}+\cdots+a_0x^{-n}\right) & = a_n\\ \lim\limits_{x\rightarrow\infty}\left(b_m + b_{m-1}x^{-1}+\cdots+b_0x^{-m}\right) & = b_m\end{aligned}\]

Therefore,

\[\lim\limits_{x\rightarrow\pm\infty} f(x) = \lim\limits_{x\rightarrow\pm\infty}x^{n-m}\left(\dfrac{a_n + a_{n-1}x^{-1}+\cdots+a_0x^{-n}}{b_m + b_{m-1}x^{-1}+\cdots+b_0x^{-m}}\right) = \frac{a_n}{b_m}\lim\limits_{x\rightarrow\pm\infty}x^{n-m}\]

THEOREM 2 Limits at Infinity of a Rational Function

The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If \(a_n,b_m\neq 0\), then

\[\lim\limits_{x\rightarrow\pm\infty}\dfrac{a_nx^n + a_{n-1}x^{n-1}+\cdots+a_0}{b_mx^m + b_{m-1}x^{m-1}+\cdots+b_0} = \frac{a_n}{b_m}\lim\limits_{x\rightarrow\pm\infty}x^{n-m}\]

Here are some examples: \[\begin{aligned} \bullet & n=m: & \lim\limits_{x\rightarrow\infty}\frac{3x^4 - 7x + 9}{7x^4 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^0 = \frac{3}{7} \\ \bullet & n < m: & \lim\limits_{x\rightarrow\infty}\frac{3x^3 - 7x + 9}{7x^4 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^{-1} = 0 \\ \bullet & n > m, n-m \text{ odd}: & \lim\limits_{x\rightarrow-\infty}\frac{3x^8 - 7x + 9}{7x^3 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow-\infty} x^5 = -\infty \\ \bullet & n > m, n-m \text{ even}: & \lim\limits_{x\rightarrow-\infty}\frac{3x^7 - 7x + 9}{7x^3 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^4 = \infty \\ \end{aligned}\]

Question 2.21 Infinite Limits Progress Check 2

Find \(\displaystyle\lim\limits_{x\rightarrow\infty}\frac{2x^4 - 3x^2 +1}{x^4 +3 x^3 + 1000x -4}\) XvVM00l89Is=
Find \(\displaystyle\lim\limits_{x\rightarrow-\infty}\frac{2x^4 - 3x^2 +1}{x^4 +3 x^3 + 1000x -4}\) XvVM00l89Is=
Find \(\displaystyle\lim\limits_{x\rightarrow\infty}\frac{2x^3 - 3x^2 +1}{x^4 +3 x^3 + 1000x -4}\) 1Wh3cvJ2xF4=
10
Correct.
Try again. Carefully determine the terms of highest degree in the numerator and denominator and work from there.
Incorrect.

Our method can be adapted to noninteger exponents and algebraic functions.

Example 3

Calculate the limits

(a) \(\displaystyle\lim\limits_{x\rightarrow \infty}\frac{3x^{7\over 2} + 7 x^{-1\over 2}}{x^2 - x^{1\over 2}}\)

(b) \(\displaystyle\lim\limits_{x\rightarrow \infty}\frac{x^2}{\sqrt{x^3+1}}\)

Solution

The Quotient Law is valid if \(\lim\limits_{x\rightarrow c}f(x) = \infty\) and \(\lim\limits_{x\rightarrow c}g(x) = L\), where \(L\neq0\):

\[\textstyle\lim\limits_{x\rightarrow c}\frac{f(x)}{g(x)} = \frac{\lim\limits_{x\rightarrow c}f(x)}{\lim\limits_{x\rightarrow c}g(x)}=\left\{\begin{aligned}\textstyle\infty,&\text{if }L>0\\-\infty,&\text{if }L<0\end{aligned}\right.\]

(a) As before, divide the numerator and denominator by \(x^2\), which is the highest power of \(x\) occurring in the denominator (this means: multiply by \(x^{-2}\)): \[\begin{aligned} \frac{3x^{7\over 2} + 7 x^{-1\over 2}}{x^2 - x^{1\over 2}} &= \left(\frac{x^{-2}}{x^{-2}}\right)\frac{3x^{7\over 2} + 7 x^{-1\over 2}}{x^2 - x^{1\over 2}} = \frac{3x^{3\over 2} + 7 x^{-5\over 2}}{1 - x^{-3\over 2}}\\ \lim\limits_{x\rightarrow\infty}\frac{3x^{7\over 2} + 7 x^{-1\over 2}}{x^2 - x^{1\over 2}} & = \frac{\lim\limits_{x\rightarrow \infty}(3x^{3\over 2} + 7 x^{-5\over 2})}{\lim\limits_{x\rightarrow\infty}(1 - x^{-3\over 2})}= \frac{\infty}{1}=\infty \end{aligned}\]

(b) The key is to observe that the denominator of \(\frac{x^2}{\sqrt{x^3+1}}\) “behaves” like \(x^{3\over2}\): \[\sqrt{x^3+1} = \sqrt{x^3(1+x^-3)} = x^{3\over2}\sqrt{1+x^{-3}}\text{ for }x > 0\] This suggests that we divide the numerator and denominator by \(x^{3\over2}\): \[\frac{x^2}{\sqrt{x^3+1}} = \left(\frac{x^{-3\over2}}{x^{-3\over2}}\right) \frac{x^2}{\sqrt{x^3+1}} = \frac{x^{1\over2}}{\sqrt{1+x^{-3}}}\] Then apply Quotient Law: \[ \lim\limits_{x\rightarrow\infty}\frac{x^2}{\sqrt{x^3+1}} = \lim\limits_{x\rightarrow\infty}\frac{x^{1\over2}}{\sqrt{1+x^{-3}}} = \frac{\lim\limits_{x\rightarrow\infty}x^{1\over2}}{\lim\limits_{x\rightarrow \infty}\sqrt{1+x^{-3}}} = \frac{\infty}{1} = \infty \]

Question 2.22 Infinite Limits Progress Check 3a

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3
Correct.
Try to factor \(x\) in the numerator and to extract \(x\) from the radical in the denominator.
Incorrect.

Question 2.23 Infinite Limits Progress Check 3b

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3
Correct.
Try again. Where is the square root function defined?
Incorrect.

Example 4

Calculate the limits at infinity of \(f(x)=\dfrac{12x+25}{\sqrt{16x^2+100x+500}}\).

Solution Divide numerator and denominator by \(x\) (multiply by \(x^{-1}\)), but notice the difference between \(x\) positive and \(x\) negative. For \(x > 0\),

\[\begin{aligned}x^{-1}\sqrt{16x^2+100x+500} & = \sqrt{x^{-2}}\sqrt{16x^2+100x+500} = \sqrt{16+{100\over x}+{500\over x^2}}\\ \lim\limits_{x\rightarrow \infty} \dfrac{12x+25}{\sqrt{16x^2+100x+500}} & = \dfrac{\lim\limits_{x\rightarrow \infty}12+{25\over x}}{\lim\limits_{x\rightarrow \infty}\sqrt{16+{100\over x}+{500\over x^2}}} = {12\over\sqrt{16}} = 3\end{aligned}\]

104

However, if \(x < 0\), then \(x=-\sqrt{x^2}\) and

\[x^{-1}\sqrt{16x^2+100x+500} = -\sqrt{x^{-2}}\sqrt{16x^2+100x+500} = -\sqrt{16+{100\over x}+{500\over x^2}}\]

So the limit as \(x\rightarrow-\infty\) is \(-3\) instead of \(3\) (Figure 2.44):

\[\lim\limits_{x\rightarrow \infty} \dfrac{12x+25}{\sqrt{16x^2+100x+500}} = \dfrac{\lim\limits_{x\rightarrow \infty}12+{25\over x}}{-\lim\limits_{x\rightarrow \infty}\sqrt{16+{100\over x}+{500\over x^2}}} = {12\over-\sqrt{16}} = -3\]

Figure 2.44: Graph of \(f(x) = \frac{12x+25}{\sqrt{16x^2+100x+500}}\) and its horizontal asymptotes \(y=-3\) and \(y=3\).

Question 2.24 Infinite Limits Progress Check 4

S+ZzbhL23HjGYZ0EuWcMeVyns26oNg4M6t82sRqa49506HpTgYT7fgGSXJwk0dRGaT+Nbjl/t3VSeRKdTC6icXfdWnp41murobXfUGG+Zv4LsO07o0j5KmyV0Lq6JLpyo8sI0ruq2flF64xgsubvqKTHfc1Gkh7d2sLwAXBiXyvbVXzQ5jW2FvlegJpCiVK0RnWQ/QlFbwNELdYpoqFs2veP+VqwbmEAiS/LVKrWc6DEtERrevODOiXn38Ts+uBRzZ/YqI2By0OzCYnWvG5CtmaophCVnaD9hLE+Bg==
3
Yes, the \(x\)-axis is the horizontal asymptote since for large values of \(x\) the denominator is approximately \(x^{6\over2}=x^3\), which has a larger exponent than the \(x^2\) in the numerator and thus \(\displaystyle \lim\limits_{x\rightarrow\infty}\dfrac{2x^2+3x+7}{\sqrt{x^6 + 2}}=0\). This function has only one horizontal asymptote since for a similar reason \(\displaystyle \lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+3x+7}{\sqrt{x^6 + 2}}=0\) also.
Try to determine what power expression \(x^q\) is a good approximation for the radical expression in the denominator. Also, make sure your answer is in the format \(y=k\)
Incorrect.

2.7.1 Section 2.7 Summary

\(\lim\limits_{x\rightarrow \infty}f(x) = L\) if \(|f(x)-L|\) becomes arbitrarily small as \(x\) increases without bound

\(\lim\limits_{x\rightarrow -\infty}f(x) = L\) if \(|f(x)-L|\) becomes arbitrarily small as \(x\) decreases without bound.