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.
Discuss the asymptotic behavior in Figure 2.41.
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
\[\lim\limits_{x\rightarrow \infty} e^x = \infty \text{ and }\lim\limits_{x\rightarrow -\infty}e^x = 0 \]
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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)
\[\lim\limits_{x\rightarrow \infty}x^n = \infty \text{ and } \lim\limits_{x\rightarrow \infty} x^{-n} = \lim\limits_{x\rightarrow \infty} \frac{1}{x^n} = 0\]
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,
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:
\[\begin{aligned}\lim\limits_{x\rightarrow \infty}(3-4x^{-3}+5x^{-5}) & = \lim\limits_{x\rightarrow \infty} 3 - 4\lim\limits_{x\rightarrow \infty}x^{-3} + 5\lim\limits_{x\rightarrow \infty}x^{-5}\\ & = 3 - 0 + 0 = 3\end{aligned}\]
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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,
\[\lim\limits_{x\rightarrow\pm\infty} f(x) = \lim\limits_{x\rightarrow\pm\infty}x^{n-m}\left(\dfrac{a_n + a_{n-1}x^{-1}+\cdots+a_0x^{-n}}{b_m + b_{m-1}x^{-1}+\cdots+b_0x^{-m}}\right) = \frac{a_n}{b_m}\lim\limits_{x\rightarrow\pm\infty}x^{n-m}\]
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: \[\begin{aligned} \bullet & n=m: & \lim\limits_{x\rightarrow\infty}\frac{3x^4 - 7x + 9}{7x^4 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^0 = \frac{3}{7} \\ \bullet & n < m: & \lim\limits_{x\rightarrow\infty}\frac{3x^3 - 7x + 9}{7x^4 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^{-1} = 0 \\ \bullet & n > m, n-m \text{ odd}: & \lim\limits_{x\rightarrow-\infty}\frac{3x^8 - 7x + 9}{7x^3 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow-\infty} x^5 = -\infty \\ \bullet & n > m, n-m \text{ even}: & \lim\limits_{x\rightarrow-\infty}\frac{3x^7 - 7x + 9}{7x^3 - 4} &= \frac{3}{7}\lim\limits_{x\rightarrow\infty} x^4 = \infty \\ \end{aligned}\]
Find \(\displaystyle\lim\limits_{x\rightarrow\infty}\frac{2x^4 - 3x^2 +1}{x^4 +3 x^3 + 1000x -4}\) | XvVM00l89Is= |
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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= |
Our method can be adapted to noninteger exponents and algebraic functions.
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: \[ \lim\limits_{x\rightarrow\infty}\frac{x^2}{\sqrt{x^3+1}} = \lim\limits_{x\rightarrow\infty}\frac{x^{1\over2}}{\sqrt{1+x^{-3}}} = \frac{\lim\limits_{x\rightarrow\infty}x^{1\over2}}{\lim\limits_{x\rightarrow \infty}\sqrt{1+x^{-3}}} = \frac{\infty}{1} = \infty \]
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 \(x > 0\),
\[\begin{aligned}x^{-1}\sqrt{16x^2+100x+500} & = \sqrt{x^{-2}}\sqrt{16x^2+100x+500} = \sqrt{16+{100\over x}+{500\over x^2}}\\ \lim\limits_{x\rightarrow \infty} \dfrac{12x+25}{\sqrt{16x^2+100x+500}} & = \dfrac{\lim\limits_{x\rightarrow \infty}12+{25\over x}}{\lim\limits_{x\rightarrow \infty}\sqrt{16+{100\over x}+{500\over x^2}}} = {12\over\sqrt{16}} = 3\end{aligned}\]
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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):
\[\lim\limits_{x\rightarrow \infty} \dfrac{12x+25}{\sqrt{16x^2+100x+500}} = \dfrac{\lim\limits_{x\rightarrow \infty}12+{25\over x}}{-\lim\limits_{x\rightarrow \infty}\sqrt{16+{100\over x}+{500\over x^2}}} = {12\over-\sqrt{16}} = -3\]
\(\lim\limits_{x\rightarrow \infty}f(x) = L\) 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.