# 4.7 | L’Hospitals Rule, Growth, and Dominance

l’Hopital’s Rule
Suppose that $$f$$ and $$g$$ are differentiable in an open interval $$I$$ that contains $$a$$, with the possible exception of $$a$$ itself, and $$g'(x) \not = 0$$ for all $$x$$ in $$I$$. If $$\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}$$ is an indeterminate form of the type $$0/0$$ or $$\infty/\infty$$, then
$$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$
provided that the limit on the right-hand side exists or is infinite.
Evaluate
$$\lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)}$$

Since the numerator and denominator both go to infinity, we have the indeterminate form $$\infty/\infty$$.

\begin{aligned} \lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} & = \lim_{x \rightarrow \infty} \frac{\frac{d}{dx}[\ln(x^8-4)]}{\frac{d}{dx}[\ln(x) \cos(1/x)]} \\ & = \lim_{x \rightarrow \infty} \frac{\frac{8x^7}{x^8-4}}{\frac{1}{x} \cos(1/x)+\frac{\ln x}{x^2} \sin(1/x)} \\ & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)(x \cos(1/x)+ \ln x \sin(1/x))} \\ & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)x \cos(1/x)} \\ & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)x} \\ & = \lim_{x \rightarrow \infty} \frac{8x^9}{x^9-4x} \\ & = 8 \end{aligned}
where

• in the first equality we applied l’Hospital’s Rule
• in the second equality we evaluated the derivatives
• in the third equality we made some algebraic simplifications
• in the fourth equality we used the fact that
$$\lim_{x \rightarrow \infty} \ln x \sin(1/x) = 0$$
• in the fifth equality we used the fact that
$$\lim_{x \rightarrow \infty} \cos(1/x) = 1$$
• in the sixth equality we made algebraic simplifications
• and for the last step, the limit at infinity of a rational function is the limit of the ratio of its leading terms

An argument using end behavior is a faster alternative to the argument above.

$$\frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} \approx \frac{\ln(x^8)}{\ln(x) \cos(0)} = \frac{8\ln(x)}{\ln(x)} = 8$$
So,
$$\lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} = \lim_{x \rightarrow \infty} 8 = 8.$$