Section LโHospitalโs Rule
Consider the following three limits.
\begin{equation*}
(1)\quad\lim_{x\to\infty} \frac{x^2 + 4x}{x} \qquad
(2)\quad\lim_{x\to\infty} \frac{x}{x^2 + 4x} \qquad
(3)\quad\lim_{x\to\infty} \frac{x^2 + 4x}{x^2}
\end{equation*}
Typically when we evaluate limits, we first attempt a direct substitution. Note that all three evaluate to the indeterminant form \(\ds\frac{\infty}{\infty}\text{.}\) Since these are all rational functions, we have some algebraic techniques we can try. Hereโs one way each can be evaluated.
\begin{align*}
(1)\quad\amp\lim_{x\to\infty} \frac{x^2 + 4x}{x} = \lim_{x\to\infty} \frac{x(x + 4)}{x}
= \lim_{x\to\infty} \left[ x+4 \right]
= \infty \\
\\
(2)\quad\amp\lim_{x\to\infty} \frac{x}{x^2 + 4x} = \lim_{x\to\infty} \frac{x}{x(x + 4)}
= \lim_{x\to\infty} \frac{1}{x + 4}
= 0 \\
\\
(3)\quad\amp\lim_{x\to\infty} \frac{x^2 + 4x}{x^2}
= \lim_{x\to\infty} \left[ 1 + \frac{4}{x}\right]
= 1 + 0
= 1
\end{align*}
Is this result surprising to you? Each had the same value, \(\ds\frac{\infty}{\infty}\text{,}\) when we did direct substitution, but each has a different final answer. Thatโs what it means to be indeterminant. we canโt tell what the answer is just based on knowing that direct substitution yields \(\ds\frac{\infty}{\infty}\text{.}\)
We can do algebra when we have rational functions, but that doesnโt work as well when we have exponential and trig functions. One technique you might consider using when you have indeterminant forms like \(\ds\frac{0}{0}\) or \(\ds\frac{\infty}{\infty}\) is LโHospitalโs Rule. Hereโs a reminder.
LโHospitalโs Rule. Suppose \(f\) and \(g\) are differentiable functions and we want to evaluate
\begin{equation*}
\lim_{x \to c} \frac{f(x)}{g(x)}
\end{equation*}
If direct substitution yields an indeterminant form \(\ds\frac{0}{0}\) or \(\pm \frac{\infty}{\infty},\) then
\begin{equation*}
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},
\end{equation*}
assuming that the limit exists.
Hereโs how it works for one of the limits above.
\begin{align*}
\lim_{x\to\infty} \frac{x^2 + 4x}{x^2}
\amp = \lim_{x\to\infty} \frac{\frac{d}{dx}\big( x^2 + 4x \big)}{\frac{d}{dx}\big(x^2\big)}
\text{ (by L'Hospital's Rule)} \\
\amp = \lim_{x\to\infty} \frac{2x + 4}{2x} \\
\amp = \lim_{x\to\infty} \frac{\frac{d}{dx}\big( 2x + 4 \big)}{\frac{d}{dx}\big( 2x \big)}
\text{ (by L'Hospital's Rule)} \\
\amp = \lim_{x\to\infty} \frac{2+0}{2} \\
\amp = \lim_{x\to\infty} 1 \\
\amp = 1
\end{align*}
Recall that we MUST check to make sure that the limit is one of those indeterminant forms before we use LโHospitalโs Rule. (And note that we did get the same answer we got before...which is good!)
If you want a more complete discussion and some examples, you can find more in your Calculus 1 book, which is available for free online (just search for "APEX calculus").
Evaluate each of the following limits. Use LโHospitalโs Rule only if it applies.
-
\(\ds \lim_{b\to \infty} \frac{b}{e^b}\)
Solution.
-
\(\ds\ds \lim_{b\to \infty} be^{-3b}\)
Solution.
-
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{5}be^{-5b} - \frac{1}{25}e^{-5b} - \frac{1}{25} \right]\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ -\frac{1}{5}be^{-5b} - \frac{1}{25}e^{-5b} - \frac{1}{25} \right] \amp = \lim_{b\to \infty} \left[ -\frac{1}{5}be^{-5b}\right] \\ \amp = -\frac{1}{5}\lim_{b\to \infty} \left[ be^{-5b}\right] \\ \amp = -\frac{1}{5}\lim_{b\to \infty} \left[ \frac{b}{e^{5b}}\right] \\ \amp = -\frac{1}{5}\lim_{b\to \infty} \left[ \frac{1}{5e^{5b}}\right] -0 - \frac{1}{25}\\ \amp = -\frac{1}{5}\cdot 0 - \frac{1}{25}\\ \amp = 0 - \frac{1}{25}\\ \amp = - \frac{1}{25} \end{align*} -
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}be^{-sb} - \frac{1}{s^2}e^{-sb} - \frac{1}{s^2} \right],\) where \(s\) is a constant and \(s \gt 0\)
Solution.
\begin{align*} \lim_{b\to \infty} \left[ -\frac{1}{s}be^{-sb} - \frac{1}{s^2}e^{-sb} - \frac{1}{s^2} \right] \amp = \lim_{b\to \infty} \left[ -\frac{1}{s}be^{-sb}\right] \\ \amp = -\frac{1}{s}\lim_{b\to \infty} \left[ \frac{b}{e^{sb}}\right] \\ \amp = -\frac{1}{s}\lim_{b\to \infty} \left[ \frac{1}{se^{sb}}\right] - \frac{1}{s^2}\cdot 0 - \frac{1}{s^2}\\ \amp = -\frac{1}{s}\cdot 0 - \frac{1}{s^2}\\ \amp = - \frac{1}{s^2} \end{align*}
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