Evaluation of $\lim_{n\rightarrow \infty}(n+\frac{1}{n})e^{\frac{1}{n}}-n$ 
Evaluation of $$\lim_{n\rightarrow \infty}\bigg(n+\frac{1}{n}\bigg)e^{\frac{1}{n}}-n$$

What i try
Let $\displaystyle \frac{1}{n}=x, $ Then $x\rightarrow 0$
So $$\lim_{x\rightarrow 0}\bigg(x+\frac{1}{x}\bigg)e^{x}-\frac{1}{x}$$
How do i solve it Help me please
 A: I work it out after your pace. Since
 $$\lim_{x\to0}\frac{e^x-1}{x}=1\\\lim_{x\to0}xe^x=0$$
We deduce that,$$\lim_{x\rightarrow 0}\bigg(x+\frac{1}{x}\bigg)e^{x}-\frac{1}{x}=1$$
A: Option:
$e^{1/n}=1+1/n+O(1/n^2)$;
$(n+1/n)e^{1/n}-n=$
$(n+1+1/n+1/n^2+O(1/n))-n=$
$1+O(1/n)$; 
A: You can also do a squeezing because the mean value theorem  and the strict monotonicity of $e^x$ give


*

*$1+\frac 1n < e^{\frac 1n} < 1+\frac{e^{\frac 1n}}{n}$
Hence,
$$\left(n+\frac 1n\right)\left(1+\frac 1n\right) -n < \left(n+\frac 1n\right)e^{\frac 1n} -n < \left(n+\frac 1n\right)\left(1+\frac{e^{\frac 1n}}{n}\right) -n$$
$$\Leftrightarrow $$
$$\underbrace{1+\frac 1n + \frac 1{n^2}}_{\stackrel{n\to\infty}{\longrightarrow}1} < \left(n+\frac 1n\right)e^{\frac 1n} -n < \underbrace{e^{\frac 1n} + \frac 1n + \frac {e^{\frac 1n}}{n^2}}_{\stackrel{n\to\infty}{\longrightarrow}1}$$
A: Continuing with you,
$\lim_{x \to 0} \frac{(x^2+1)e^x-1}{x}=lim_{x \to 0}\frac{(2x)e^x+(x^2+1)e^x}{1}$ [Differentiating both numerator and denominator using L'Hospital rule,since it is a $\frac{0}{0}$ form.]
Now,it is clear that the limit$=(2*0)*e^0+(0^2+1)*e^0=1.$
