Question

If $\underset{x\to 0}{lim}{(1+x(1+\frac{f\left(x\right)}{k{x}^2}))}^{1/x}=e^3$ and $f\left(4\right)=64$,

then the value of $k$ has is?

• A. $1$
• B. $2$
• C. $4$
• D. none of these

Answer 1

The correct option is B $2$

$\underset{x\to 0}{lim}{(1+x(1+\frac{f\left(x\right)}{k{x}^2}))}^{\frac{1}{x}}=e^3$
Taking log on both sides
$\underset{x\to 0}{lim}\frac{log(1+x(1+\frac{f\left(x\right)}{k{x}^2}))}{x}=3$
Now expanding the log term upto its first term
$\underset{x\to 0}{lim}(1+\frac{f\left(x\right)}{k{x}^2})=3$
$f\left(x\right)=2k{x}^2$
Given
$f\left(4\right)=64$
$\Rightarrow k=2$