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Section B.1 Limits at infinity

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In this section, we’ll 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".
  1. Evaluate each of the following limits.
    1. \(\ds \lim_{b\to \infty} \frac{1}{12}\)
      Solution.
      The limit of a constant is the constant, so
      \begin{equation*} \lim_{b\to \infty} \frac{1}{12} = \frac{1}{12}. \end{equation*}
      Answer.
      \(\frac{1}{12} \)
    2. \(\ds \lim_{b\to \infty} \frac{1}{s},\) where \(s\) is a constant
      Solution.
      The limit of a constant is the constant, so
      \begin{equation*} \lim_{b\to \infty} \frac{1}{s} = \frac{1}{s}. \end{equation*}
      Answer.
      \(\frac{1}{s} \)
    3. \(\ds \lim_{b\to \infty} e^{2b}\)
      Solution.
      As \(b\) increases to large positive values, the value of \(e^{2b}\) also increases to large positive values. Therefore
      \begin{equation*} \lim_{b\to \infty} e^{2b} = \infty. \end{equation*}
      Answer.
      \(\infty \)
    4. \(\ds \lim_{b\to \infty} e^{0.1b}\)
      Solution.
      As \(b\) increases to large positive values, the value of \(e^{0.1b}\) also increases to large positive values. Therefore
      \begin{equation*} \lim_{b\to \infty} e^{0.1b} = \infty. \end{equation*}
      Answer.
      \(\infty \)
    5. \(\ds \lim_{b\to \infty} e^{-2b}\)
      Solution.
      As \(b\) increases to large positive values, the value of \(e^{-2b}\) decreases to zero. Therefore
      \begin{equation*} \lim_{b\to \infty} e^{-2b} = 0. \end{equation*}
      Answer.
      \(0 \)
  2. 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, you’ll come to realize that:
    • If \(a=0,\) then \(e^{ab} = e^{0} = 1,\) so \(\ds \lim_{b \to \infty}e^{ab} = 1\text{.}\)
    • 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 \(a \le 0\text{.}\)
    Answer.
    \(a \le 0 \)
  3. Evaluate each of the following limits.
    1. \(\ds \lim_{b\to \infty} \left[ -\frac{1}{-3}e^{-3b} + \frac{1}{-3} \right]\)
      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*}
      Answer.
      \(\ds -\frac{1}{3} \)
    2. \(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right]\) 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*}
      Answer.
      \(\ds \frac{1}{s} \)
    3. \(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s} \right]\) 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*}
      Answer.
      \(\infty \)
    4. \(\ds \lim_{b\to \infty} \left[ \frac{1}{s^2}e^{-sb} \right]\) where \(s\) is a constant and \(s \gt 0\text{.}\)
      Solution.
      \begin{align*} \lim_{b\to \infty} \left[ \frac{1}{s^2}e^{-sb} \right] \amp = \frac{1}{s^2}\lim_{b\to \infty} \left[ e^{-sb} \right]\\ \amp = \frac{1}{s^2}\cdot 0 \\ \amp = 0 \end{align*}
      Answer.
      \(0 \)
    5. \(\ds \lim_{b\to\infty} \left[\frac{1}{s-3}e^{(3-s)b} + \frac{1}{3-s}\right]\) 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*}
      Answer.
      \(\ds \frac{1}{3-s} \)
    6. \(\ds \lim_{b\to \infty} \left[ \frac{1}{s+7}e^{(-7-s)b} + \frac{1}{-7-s} \right]\) 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*}
      Answer.
      \(\ds \frac{1}{-7-s} \)
    7. \(\ds \lim_{b\to \infty} \left[ \frac{1}{s-a}e^{(a-s)b} + \frac{1}{a-s} \right]\) 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*}
      Answer.
      \(\ds \frac{1}{a-s} \)
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