## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 5

#### Answer

Using the sequence $$\lim_{x\to a}f(x)g(x)=\lim_{x\to a}\frac{f(x)}{\frac{1}{g(x)}}$$ or $$\lim_{x\to a}f(x)g(x)=\lim_{x\to a}\frac{g(x)}{\frac{1}{f(x)}}.$$

#### Work Step by Step

Suppose that $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=\pm\infty.$ Then using the sequence $$\lim_{x\to a}f(x)g(x)=\lim_{x\to a}\frac{f(x)}{\frac{1}{g(x)}}$$ and noting that $1/g(x)$ goes to $1/\infty$ i.e. to $0$ when $x\to a$ we get the expression of the form of $0/0$. Also using the sequence $$\lim_{x\to a}f(x)g(x)=\lim_{x\to a}\frac{g(x)}{\frac{1}{f(x)}},$$ and noting that $1/f(x)$ goes to $1/0^{\pm}$ i.e. to $\pm\infty$ (it is needed that $f$ approaches zero from one side to do this transformation) so we get the form $\infty/\infty$.

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