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Section B.2 L’Hospital’s 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 indeterminate form \(\ds\frac{\infty}{\infty}\text{.}\) Since these are all rational functions, we have some algebraic techniques we can try. Here’s 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. That’s what it means to be indeterminate. We can’t 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 doesn’t work as well when we have exponential and trig functions. One technique you might consider using when you have indeterminate forms like
\(\ds\frac{0}{0}\) or
\(\ds\frac{\infty}{\infty}\) is L’Hospital’s Rule. Here’s a reminder.
✳️ L’Hospital’s 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 indeterminate 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.
Here’s 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 indeterminate forms before we use L’Hospital’s Rule. (And note that we did get the same answer we got before...which is good!)
Exercises Exercises
Evaluate each of the following limits. Use L’Hospital’s Rule only if it applies.
1.
\(\ds \lim_{b\to \infty} \frac{b}{e^b}\)
2.
\(\ds\ds \lim_{b\to \infty} be^{-3b}\)
3.
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{5}be^{-5b} - \frac{1}{25}e^{-5b} - \frac{1}{25} \right]\)
4.
\(\ds \lim_{b\to \infty} \left[ -\frac{1}{s}be^{-sb} - \frac{1}{s^2}e^{-sb} - \frac{1}{s^2} \right],\) where
\(s\) is a constant and
\(s \gt 0\)
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