Question

$\underset{x\to 0}{lim}\frac{e^x+\sin x-1}{\log (1+x)}$=

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $2$

Answer 1

The correct option is D $2$

$\underset{x\to 0}{lim}\frac{e^x+\sin x-1}{\log (1+x)}$
$=\underset{x\to 0}{lim}\frac{(1+x+\frac{x^2}{2!}+....)+(x-\frac{x^3}{3!}+.....)-1}{x-\frac{1}{2}{x}^2+.......}$
$=\underset{x\to 0}{lim}\frac{x(1+\frac{x}{2!}+....)+x(1-\frac{x^2}{3!}+.....)}{x(1-\frac{1}{2}x+.......)}=\frac{(1+0)+(1-0)}{1-0}=2$