Question

Evaluate:$\underset{x\to 0}{\lim }\frac{e^x-e^{\sin x}}{x-\sin x}$

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C

$1$

Explanation for the correct answer:

Simplifying the equation to determinate form and applying the limits:

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left(\frac{e^x-e^{\sin x}}{x-\sin x}\right) \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{e^x-e^{\sin x}\cos x}{1-\cos x}\right)\left[\text{Differentiating}\right] \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{e^x-\left(e^{\sin x}\left(-\sin x\right)+e^{\sin x}\cos x·\cos x\right)}{0-\left(-\sin x\right)}\right)\left[\text{Differentiating}\right] \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{e^x+e^{\sin x}.\sin x-{\cos }^{2}x·x{e}^{\sin x}}{\sin x}\right) \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{e^x+e^{\sin x}\cos x+\sin x·e^{\sin x}-{\cos }^{2}x·e^{\sin x}{\cos }^{x-2\cos x\left(-\sin x\right)e^{\sin x}}}{\cos x}\right)\left[\text{Differentiating}\right] \end{gathered}$$

Applying the limits

$$\begin{gathered} \Rightarrow \frac{e^0+e^0\times 1+0·e^0·1-1^2·e^0·1-2·1\left(-0\right)e^0}{1} \\ \Rightarrow 1+1-1 \\ \Rightarrow 1 \end{gathered}$$

Thus, $\underset{x\to 0}{\lim }\frac{e^x-e^{\sin x}}{x-\sin x}=1$

Therefore, the correct answer is option (C).