Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

Answer: (C)

$2$

Now,

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

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

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

$=\frac{1+1}{1}$

$=2$.