In this section, you will learn what L'Hopital's rule is, and how to use it. You will see that some problems require using L'Hopital's rule more than once. Before attempting this section, you should be excellent at finding derivatives.
L'Hopital's rule
Typeset step-by-step solutions.
\(\displaystyle \lim_{x \to 0}\left(\dfrac{e^x - 1}{x}\right)\)
\(1\)
\(\displaystyle \lim_{x \to 0}\left(\dfrac{e^x - 1}{x^3}\right)\)
\(\infty\)
\(\displaystyle \lim_{x \to 0}\left(\dfrac{2x^2}{\cos x - 1}\right)\)
\(-4\)
\(\displaystyle \lim_{x \to 1}\left(\dfrac{x^9 - 1}{x^5 - 1}\right)\)
\(\dfrac{9}{5}\)
\(\displaystyle\lim_{x \to \infty}\dfrac{x^2}{e^x}\)
\(0\)
\(\displaystyle\lim_{x \to \infty}\dfrac{\ln x}{x}\)
\(0\)
\(\displaystyle\lim_{x \to 0}\dfrac{\sin(7x)}{\sin(4x)}\)
\(\dfrac{7}{4}\)
\(\displaystyle\lim_{x \to 0}\dfrac{\sin(8x)}{3x}\)
\(\dfrac{8}{3}\)
\(\displaystyle\lim_{x \to \infty}x\ln x\)
\(\infty\)
\(\displaystyle\lim_{x \to \infty}x^{1 / x}\)
\(1\)
\(\displaystyle\lim_{x \to 0}(1 - 2x)^{1 / x}\)
\(e^{-2}\)
\(\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x}\)
\(1\)
