Question

The value of the limit $\underset{x\to 1}{lim}\frac{\sin (e^{x-1}-1)}{\log x}$ is

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

Answer 1

The correct option is D $1$

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