Question

Evaluate :$\underset{x\to 0}{\lim }\frac{{\mathrm{\pi }}^{\mathrm{x}}-1}{\sqrt{1+x}-1}$

• A. does not exist
• B. is equal to ${\log }_e\left({\mathrm{\pi }}^2\right)$
• C. is equal to $1$
• D. lies between $10$ and $11$

Answer 1

The correct option is B

is equal to ${\log }_e\left({\mathrm{\pi }}^2\right)$

Explanation for the correct option:

Find the value of $\underset{x\to 0}{\lim }\frac{{\mathrm{\pi }}^{\mathrm{x}}-1}{\sqrt{1+x}-1}$

Consider given value as

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\mathrm{\pi }}^{\mathrm{x}}-1}{\sqrt{1+x}-1} \\ \Rightarrow I=\frac{{\mathrm{\pi }}^0-1}{\sqrt{1+0}-1} \\ \Rightarrow I=\frac{{\mathrm{\pi }}^0-1}{1-1} \\ \Rightarrow I=\frac{0}{0}\left[\text{indeterminent Form}\right] \end{gathered}$$

Using the L' hospital rule

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\mathrm{\pi }}^{\mathrm{x}}-1}{\sqrt{1+x}-1} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\mathrm{\pi }}^{\mathrm{x}}\log \mathrm{\pi }}{{\displaystyle \frac{1}{2\sqrt{1+x}}-0}} \\ \Rightarrow I=\frac{{\mathrm{\pi }}^0\log \mathrm{\pi }}{{\displaystyle \frac{1}{2\sqrt{1+0}}-1}} \\ \Rightarrow I=\frac{2\log \mathrm{\pi }}{{\displaystyle 1}} \\ \Rightarrow I=2\log \mathrm{\pi } \\ \Rightarrow I=\log {\mathrm{\pi }}^2 \end{gathered}$$

Hence, the correct answer is option B.