# Evaluation of $\lim\limits_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]$

Evaluation of $\displaystyle \lim_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]$

$\bf{My\; Try::}$ Let $\displaystyle l=\displaystyle \lim_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]=\lim_{x\rightarrow \infty}x\left[1+(x-1)\ln\left(1+\frac{1}{x}\right)\right]$

So we get $$l=\lim_{x\rightarrow \infty}x\left[1+(1-x)\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}-.........\infty\right)\right]$$

So we get $$l=\lim_{x\rightarrow \infty}x\left[1-1+\frac{1}{2x}-\frac{1}{3x^2}+\frac{1}{4x^3}+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}.....\right] = \frac{3}{2}$$

My Question is How can we solve it without using series expansion,

If yes then plz explain here, Thanks

Let $y = 1/x$ and apply L'Hospital's rule:
\begin{align}l &= \lim_{y \to 0}\frac{1 + (1 - 1/y)\ln(1+y)}{y}\\ &= \lim_{y \to 0}\frac{1 - \frac{\ln(1+y)}{y}}{y} + \lim_{y \to 0}\frac{\ln(1+y)}{y} \\ &= \lim_{y \to 0}\frac{\ln(1+y)(1+y)-y}{y^2(1+y)}+ \lim_{y \to 0}\frac{1}{1+y} \\ &= \lim_{y \to 0}\frac{1 + \ln(1+y) - 1}{2y + 3y^2}{}+1 \\ &= \lim_{y \to 0}\frac{1 }{(1+y)(2 + 6y)}{}+1\\ &= \frac{1}{2} + 1 \\ &= \frac{3}{2}\end{align}
At first i changed variable, then i developed the $\ln$ with Taylor formula $$\lim_{x\rightarrow \infty}x\left[\ln \left(e\left(1+\frac{1}{x}\right)^{1-x}\right)\right]=\lim _{t\to 0}\left(\frac{\left[\ln \left(e\left(1+t\right)^{1-\frac{1}{t}}\right)\right]}{t}\right) = \lim _{t\to 0}\left(\frac{\left[1+\ln \:\left(t+1\right)\left(-\frac{1}{t}+1\right)\right]}{t}\right) \approx \lim _{t\to 0}\left(\frac{\left[1+\left(t-\frac{1}{2}t^2+o\left(t^2\right)\right)\left(-\frac{1}{t}+1\right)\right]}{t}\right) = \color{red}{\frac{3}{2}}$$