Question

The value of $\underset{x\to 2}{\lim }\frac{\left[e^{3x-6}-1\right]}{\sin \left(2-x\right)}$ is

• A. $\frac{3}{2}$
• B. $3$
• C. $-3$
• D. $-1$

Answer 1

The correct option is C

$-3$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 2}{\lim }\frac{\left[e^{3x-6}-1\right]}{\sin \left(2-x\right)}$

$$\begin{gathered} =\frac{e^{6-6}-1}{\sin \left(2-2\right)} \\ =\frac{1-1}{\sin \left(0\right)} \\ =\frac{0}{0} \end{gathered}$$

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

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{e^{3x-6}\left(3\right)}{\cos \left(2-x\right)(0-1)} \\ =-\frac{e^{6-6}\left(3\right)}{\cos \left(0\right)} \\ =-3 \end{gathered}$$

Therefore, the value of $\underset{x\to 2}{\lim }\frac{\left[e^{3x-6}-1\right]}{\sin \left(2-x\right)}$ is $-3$.

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