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Section6.5L'Hôpital's Rule

Limits involving combinations of continuous functions are generally computed by evaluating the values of the functions. Arithmetic involving infinity will follow elementary rules. Suppose \(L \gt 0\) is a positive number. Then when performing arithmetic, the following rules will hold. \begin{gather*} \infty \pm L = \infty \\ \infty + \infty = \infty \\ -\infty + -\infty = -\infty \\ L \cdot \infty = \infty \\ -L \cdot \infty = -\infty \\ \infty \cdot \infty = \infty \\ \infty \cdot -\infty = -\infty \\ -\infty \cdot -\infty = \infty \end{gather*}

However, when calculations would result that attempt to cancel away an infinite quantity (or zero), the limit has an indeterminate form. That is, calculations involving any of the following arithmetic are in an indeterminate form, \begin{equation*}\frac{\infty}{\infty} \qquad \infty - \infty \qquad \frac{0}{0} \qquad 0 \cdot \infty.\end{equation*} The general strategy for evaluating indeterminate limits involves rewriting the limit in a different form, or finding another limit that is known to be equivalent.

L'Hôpital's rule is a theorem that allows us to rewrite a limit which has an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) using derivatives.

Proof

It is essential that you verify that the hypotheses of L'Hôpital's rule apply before replacing the derivative. Limits and derivatives are not the same. Computing a limit does not always mean we are computing a derivative.

In addition, make note that changing the limit using L'Hôpital's rule is not computing the derivative of a quotient. It is replacing the limit of a quotient with the limit of a new quotient involving two derivatives.

Our first few examples will illustrate that L'Hôpital's rule gives the same result as methods we learned earlier involving factoring.

Example6.5.2

Evaluate the limit \(\displaystyle \lim_{x \to 2} \frac{x^2-3x+2}{x^2+x-6}\) using factoring and using L'Hôpital's rule.

Solution

Sometimes, we need to apply L'Hôpital's rule more than once when the modified limit is still in indeterminate form.

Example6.5.3

Evaluate the limit \(\displaystyle \lim_{x \to \infty} \frac{2x^2+x-3}{x^2-x-5}\) using factoring and using L'Hôpital's rule.

Solution

One of the advantages of L'Hôpital's rule is that it allows us to evaluate limits where factoring does not help.

Example6.5.4

Compute \(\displaystyle \lim_{x \to 3} \frac{2^x-8}{x^2+x-12}\).

Solution

Indeterminate limits that are not fractions of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) do not directly apply L'Hôpital's rule. You must first use algebra to rewrite them in a way that they do have the appropriate form.

Example6.5.5

Evaluate \(\displaystyle \lim_{x \to 0^+} x \ln(x)\).

Solution
Example6.5.6

Evaluate \(\displaystyle \lim_{x \to \infty} x^2e^{-3x}\).

Solution

We end with an example involving powers. When both the base and the exponent are variable, we must interpret a power in terms of composition with the exponential function, \begin{equation*} u^v = \exp( \ln(u^v)) = \exp(v \ln(u)) = e^{v \ln(u)}.\end{equation*} Because the natural exponential function is continuous, we only need to evaluate the limit of \(v \ln(u)\) and then evaluate the exponential function at the corresponding limit. This is a consequence of Theorem 5.7.11.

Example6.5.7

Evaluate \(\displaystyle \lim_{x \to \infty} (1 + \frac{r}{x})^{xt}\), where \(r\) and \(t\) are constant values.

Solution