Question

If $\underset{x\to 0}{lim}{[1+x+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}=e^3$, then the value of $ln\left(\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}\right)$ is .............

• A. $3$
• B. $2$
• C. $-2$
• D. $4$

Answer 1

Answer: (B)

$2$

$\underset{x\to 0}{lim}{[1+x+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}=e^3$
or $\underset{x\to 0}{lim}{e}^{{lim}_{x\to 0}{[1+x+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}}=e^3$
or $\underset{x\to 0}{lim}{e}^{{lim}_{x\to 0}[1+\frac{f\left(x\right)}{x^2}]}=e^3$
or $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}=2$
Now, $\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}=e^{{lim}_{x\to 0}{[1+\frac{f\left(x\right)}{x}-1]}^{\frac{1}{x}}}=e^{{lim}_{x\to 0}\frac{f\left(x\right)}{x^2}}=e^2$

$\Rightarrow \underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x}]}^{\frac{1}{x}}=2$

Hence, option 'B' is correct.