Question

$\underset{x\to 0}{\lim }\frac{e^x+e^{-x}+2\cos x-4}{x^4}$ is equal to

• A. $0$
• B. $1$
• C. $\frac{1}{6}$
• D. $-\frac{1}{6}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{e^x+e^{-x}+2\cos x-4}{x^4}$
$(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}-2\sin x}{4x^3}$
$(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2\cos x}{12x^2}$
$(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{e^x-e^{-x}+2\sin x}{24x}$
$(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{e^x+e^{-x}+2\cos x}{24}$
$=\frac{4}{24}=\frac{1}{6}$