Question

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

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

Answer 1

The correct option is A $0$

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

Substitute $x=0$

$\frac{e^0+e^{-0}+2\cos 0-4}{0^2}=\frac{1+1+2-4}{0}=\frac{0}{0}$

This is an indeterminate type so use L-Hospital's Rule which is the limit of the quotient of the derivative of the top and the derivative of the bottom as $x$ goes to $0$.

$=\frac{\frac{d}{dx}(e^x+e^{-x}+2\cos x-4)}{\frac{d}{dx}{x}^2}$

$=\frac{(e^x-e^{-x}-2\sin x)}{2x}$

Substitute $x=0$

$=\frac{(e^0-e^{-0}-2\sin 0)}{2\left(0\right)}=\frac{1-1-0}{0}=\frac{0}{0}$

Apply L-Hopital's Rule, as this is still indeterminate.

$=\frac{\frac{d}{dx}(e^x-e^{-x}-2\sin x)}{\frac{d}{dx}2x}$

$=\frac{(e^x+e^{-x}-2\cos x)}{2}$

Substitute $x=0$

$=\frac{(e^0+e^{-0}-2\cos 0)}{2}=\frac{1+1-2}{2}=0$