Question

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

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

Answer 1

Answer: (C)

$\frac{1}{6}$

Let, $L=\underset{x\to 0}{lim}\frac{e^x+e^{-x}+2cosx-4}{x^4}$
Using series expansion,
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+................$
$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+................$
and, $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..............$
$\therefore L=\underset{x\to 0}{lim}\frac{2(2+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...........)+2(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...........)-4}{x^4}$
$=2(\frac{1}{4!}+\frac{1}{4!})=\frac{1}{6}$