Section Limits at infinity
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In this section, weβll focus our attention on limits at infinity. If you want a refresher, check out your Calculus 1 book (available online--just search for "APEX calculus".
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Evaluate each of the following limits.
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\(\ds \lim_{b\to \infty} \frac{1}{12}\)
Solution.
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\(\ds \lim_{b\to \infty} \frac{1}{s},\) where \(s\) is a constant
Solution.
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\(\ds \lim_{b\to \infty} e^{2b}\)
Solution.
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\(\ds \lim_{b\to \infty} e^{0.1b}\)
Solution.
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\(\ds \lim_{b\to \infty} e^{-2b}\)
Solution.
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For what values of \(a\) is the limit \(\ds \lim_{b \to \infty}e^{ab}\) finite?
Solution.
Hopefully you thought a little bit about this as you answered the previous question. If you think about it for a little bit, youβll come to realize that:-
If \(a=0,\) then \(e^{ab} = e^{0} = 1,\) so \(\ds \lim_{b \to \infty}e^{ab} = 1\text{.}\)
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If \(a\) is negative, then as \(b\) increases to large positive values, the value of \(e^{ab}\) decreases to zero. Therefore \(\ds \lim_{b\to \infty} e^{ab} = 0.\)
Hence, the limit \(\ds \lim_{b \to \infty}e^{ab}\) finite when \(a \le 0\text{.}\) -
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Evaluate each of the following limits.
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\(\ds \lim_{b\to \infty} \left[ -\frac{1}{-3}e^{-3b} + \frac{1}{-3} \right]\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ -\frac{1}{-3}e^{-3b} + \frac{1}{-3} \right] \amp = -\frac{1}{-3}\lim_{b\to \infty} \left[ e^{-3b}\right] + \lim_{b\to \infty} \left[\frac{1}{-3} \right] \\ \amp = -\frac{1}{-3}\cdot 0 + \frac{1}{-3} \\ \amp = 0 - \frac{1}{3} \\ \amp = -\frac{1}{3} \end{align*} -
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right]\) where \(s\) is a constant and \(s \gt 0\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right] \amp = \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb}\right] + \lim_{b\to \infty} \left[\frac{1}{s} \right] \\ \amp = -\frac{1}{s}\lim_{b\to \infty} \left[ e^{-sb}\right] + \frac{1}{s} \\ \amp = -\frac{1}{s}\cdot 0 + \frac{1}{s} \\ \amp = \frac{1}{s} \end{align*} -
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right]\) where \(s\) is a constant and \(s \lt 0\text{.}\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right] \amp = \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb}\right] + \lim_{b\to \infty} \left[\frac{1}{s} \right] \\ \amp = -\frac{1}{s}\lim_{b\to \infty} \left[ e^{-sb}\right] + \frac{1}{s} \\ \amp = -\frac{1}{s}\cdot \infty + \frac{1}{s} \\ \amp = \infty + \frac{1}{s} \\ \amp = \infty \end{align*} -
\(\ds \lim_{b\to \infty} \left[ \frac{1}{s^2}e^{-sb} \right]\) where \(s\) is a constant and \(s \gt 0\text{.}\)
Solution.
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\(\ds \lim_{b\to\infty} \left[\frac{1}{s-3}e^{(3-s)b} + \frac{1}{3-s}\right]\) where \(s\) is a constant and \(s>3\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ \frac{1}{s-3}e^{(3-s)b} + \frac{1}{3-s} \right] \amp = \lim_{b\to \infty} \left[ \frac{1}{s-3}e^{(3-s)b}\right] + \lim_{b\to \infty} \left[ \frac{1}{3-s} \right] \\ \amp = \frac{1}{s-3}\lim_{b\to \infty} \left[ e^{(3-s)b}\right] + \frac{1}{3-s} \\ \amp = \frac{1}{s-3}\cdot 0+ \frac{1}{3-s} \\ \amp = \frac{1}{3-s} \end{align*} -
\(\ds \lim_{b\to \infty} \left[ \frac{1}{s+7}e^{(-7-s)b} + \frac{1}{-7-s} \right]\) where \(s\) is a constant and \(s \gt -7\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ \frac{1}{s+7}e^{(-7-s)b} + \frac{1}{-7-s} \right] \amp \lim_{b\to \infty} \left[ \frac{1}{s+7}e^{(-7-s)b}\right] + \lim_{b\to \infty} \left[\frac{1}{-7-s} \right] \\ \amp \frac{1}{s+7}\lim_{b\to \infty} \left[e^{(-7-s)b}\right] + \frac{1}{-7-s} \\ \amp = \frac{1}{s+7}\cdot 0 + \frac{1}{-7-s} \\ \amp = \frac{1}{-7-s} \end{align*} -
\(\ds \lim_{b\to \infty} \left[ \frac{1}{s-a}e^{(a-s)b} + \frac{1}{a-s} \right]\) where \(s\) is a constant and \(s \lt a\text{.}\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ \frac{1}{s-a}e^{(a-s)b} + \frac{1}{a-s} \right] \amp \lim_{b\to \infty} \left[ \frac{1}{s-a}e^{(a-s)b} \right] + \lim_{b\to \infty} \left[\frac{1}{a-s} \right] \\ \amp \frac{1}{s-a}\lim_{b\to \infty} \left[ e^{(a-s)b} \right] + \frac{1}{a-s} \\ \amp \frac{1}{s-a}\cdot 0 + \frac{1}{a-s} \\ \amp = \frac{1}{a-s} \end{align*}
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