Question

Evaluate:

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

• A. $1$
• B. $2$
• C. $-1$
• D. none of these

Answer 1

The correct option is A $1$

Now,

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

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

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

$=\frac{1.1}{1^2}$

$=1$.