Question

Let $f\left(x\right)$ is defined as ${(1+\frac{1}{x})}^{f\left(x\right)+x}=e,$ then ${lim}_{x\to \infty }f\left(x\right)=\frac{1}{k}.$ Then the value of $k$ is

Answer 1

Let, ${(1+\frac{1}{x})}^{f\left(x\right)+x}=e$

Taking natural logarithm on both sides, we get

$\left(f\right(x)+x)\log (1+\frac{1}{x})=1$

On rearranging,

$f\left(x\right)=\frac{1}{\log \left(\frac{x+1}{x}\right)}-x$

Let $\frac{x+1}{x}=t$

As, $x\to \infty ,t\to 1$, and $x=\frac{1}{t-1}$

$\therefore \underset{x\to \infty }{lim}f\left(x\right)=\underset{t\to 1}{lim}\frac{1}{\log t}-\frac{1}{t-1}$

$\underset{x\to \infty }{lim}f\left(x\right)=\underset{t\to 1}{lim}\frac{t-1-\log t}{(t-1)\log t}=\underset{t\to 0}{lim}\frac{t-\log (t+1)}{t\log (t+1)}$

Using expansion of $\log (t+1)=t-\frac{t^2}{2}+\frac{t^3}{3}.......,$ we get

$\underset{x\to \infty }{lim}f\left(x\right)=\underset{t\to 0}{lim}\frac{\frac{1}{2}{t}^2-\frac{1}{3}{t}^3+.....}{t^2-\frac{t^3}{3}-.....}=\frac{1}{2}=\frac{1}{k}$

$\therefore k=2$