Question

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

• A. $1$
• B. $0$
• C. $e$
• D. $e^{-1}$

Answer 1

Answer: (A)

$1$

$\underset{x\to 1}{lim}\frac{\sin (e^{x-1}-1)}{\log x}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule
$=\underset{x\to 1}{lim}\frac{e^{x-1}\cos (e^{x-1}-1)}{1/x}$
$=\frac{e^0\cos 0}{1}=1$