Question

$\underset{x\to 0}{\lim }{(1-ax)}^{\frac{1}{x}}=$

• A. $e^{-a}$
• B. $e$
• C. $e^a$
• D. $1$

Answer 1

The correct option is A

$e^{-a}$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{x\to 0}{\lim }{(1-ax)}^{\frac{1}{x}} \\ =e^{\left(\ln \left(\underset{x\to 0}{\lim }(1-ax)\right)\right)} \\ =e^{\left(\underset{x\to 0}{\lim }\frac{\ln \left(1-ax\right)}{x}\right)} \\ =e^{\left(\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d}{dx}\ln \left(1-ax\right)}}{{\displaystyle \frac{d}{dx}\left(x\right)}}\right)}\left[\mathrm{We}\mathrm{have}\mathrm{indeterminate}\mathrm{form}\frac{0}{0}\mathrm{L}\text{'}\mathrm{Hospital}\text{'}\mathrm{s}\mathrm{rule}\mathrm{applies}\right] \\ =e^{\left(\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{-a}{1-ax}}}{1}\right)} \end{gathered}$$

Applying the limits;

$$\begin{gathered} =e^{\left(\frac{{\displaystyle \frac{-a}{1-0}}}{1}\right)} \\ =e^{-a} \end{gathered}$$

Therefore, the correct answer is option (A).