Question

The value of $\underset{x\to 1}{\lim }\frac{\left[\sin \left(e^{x-1}-1\right)\right]}{\ln \left(x\right)}$ is

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

Answer 1

The correct option is D

$1$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 1}{\lim }\frac{\left[\sin \left(e^{x-1}-1\right)\right]}{\ln \left(x\right)}$

$$$$\begin{gathered} =\frac{\left[\sin \left(e^{1-1}-1\right)\right]}{\ln \left(1\right)} \\ =\frac{0}{0} \end{gathered}$$

Since, it is $\frac{0}{0}$ form,.apply L.Hospital's rule,

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\cos \left(e^{x-1}-1\right)\left(e^{x-1}\right)}{{\displaystyle \frac{1}{x}}} \\ =\frac{\cos \left(e^{1-1}-1\right)\left(e^{1-1}\right)}{{\displaystyle \frac{1}{1}}} \\ =\frac{\cos \left(0\right)\times 1}{1}\left[\cos \left(0\right)=1\right] \\ =1 \end{gathered}$$

Therefore, the value of $\underset{x\to 1}{\lim }\frac{\left[\sin \left(e^{x-1}-1\right)\right]}{\ln \left(x\right)}$ is $1$.

Hence, option $\left(D\right)$ is the correct option.