The value expression \[\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\sin {\text{x}}}}{{{{\text{e}}^{\text{x}}}{\text{x}}}}\] is A. 0 B. \[\frac{1}{2}\] C. 1 D. \[\frac{1}{{1 + {\text{e}}}}\]

”0″
”[ rac{1}{2}\
” option3=”1″ option4=”\[\frac{1}{{1 + {\text{e}}}}\]” correct=”option3″]

The correct answer is $\frac{1}{{1 + {\text{e}}}}$.

We can evaluate the limit using L’Hôpital’s rule, which states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists and $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ also exists, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$.

In this case, we have $f(x)=\sin x$ and $g(x)=e^x$. We can differentiate both $f(x)$ and $g(x)$ to get $f'(x)=\cos x$ and $g'(x)=e^x$. Substituting these into L’Hôpital’s rule, we get

$$\begin{align}
\lim_{x\to 0} \frac{{\sin {\text{x}}}}{{{{\text{e}}^{\text{x}}}{\text{x}}}} &= \lim_{x\to 0} \frac{{f'(x)}}{{g'(x)}} \\
&= \lim_{x\to 0} \frac{{\cos {\text{x}}}}{{{{\text{e}}^{\text{x}}}} \\
&= \frac{1}{{1 + {\text{e}}}}
\end{align
}$$

Therefore, the value of the expression $\lim_{x\to 0} \frac{{\sin {\text{x}}}}{{{{\text{e}}^{\text{x}}}{\text{x}}}}$ is $\frac{1}{{1 + {\text{e}}}}$.