Question

Evaluate:$\underset{x\to 1}{\lim }\left(\frac{e^{-x}-e^{-1}}{x-1}\right)$

• A. $\frac{1}{e}$
• B. $-\frac{1}{e}$
• C. $1$
• D. None of these

Answer 1

The correct option is B

$-\frac{1}{e}$

Explanation for the correct option:

Find the value of $\underset{x\to 1}{\lim }\left(\frac{e^{-x}-e^{-1}}{x-1}\right)$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{e^{-x}-e^{-1}}{x-1}\right) \\ I=\left(\frac{e^{-1}-e^{-1}}{1-1}\right) \\ I=\frac{0}{0}\left[\text{indeterminate form}\right] \end{gathered}$$

Using the L' hospital rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

Then,

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

Hence, the correct answer is option B.