# 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

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,$$\lim_{x\rightarrow 0}\bigg(x+\frac{1}{x}\bigg)e^{x}-\frac{1}{x}=1$$

Option:

$$e^{1/n}=1+1/n+O(1/n^2)$$;

$$(n+1/n)e^{1/n}-n=$$

$$(n+1+1/n+1/n^2+O(1/n))-n=$$

$$1+O(1/n)$$;

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,

$$\left(n+\frac 1n\right)\left(1+\frac 1n\right) -n < \left(n+\frac 1n\right)e^{\frac 1n} -n < \left(n+\frac 1n\right)\left(1+\frac{e^{\frac 1n}}{n}\right) -n$$ $$\Leftrightarrow$$ $$\underbrace{1+\frac 1n + \frac 1{n^2}}_{\stackrel{n\to\infty}{\longrightarrow}1} < \left(n+\frac 1n\right)e^{\frac 1n} -n < \underbrace{e^{\frac 1n} + \frac 1n + \frac {e^{\frac 1n}}{n^2}}_{\stackrel{n\to\infty}{\longrightarrow}1}$$

• Thanks Trancelocation please explain me how i prove $e^{x}<1+xe^{x}$ for $x>0$ Jan 1, 2020 at 9:17
• @jacky : The mean value theorem gives for $x > 0$ $$e^x-e^0 = e^{\xi}(x-0) \mbox{ with } 0 < \xi < x$$ Hence, using the strict monotonicity of $e^x$ you get $$xe^0 < e^x-1 < xe^x$$ $$\Leftrightarrow 1+x < e^x < 1+xe^x$$ Jan 1, 2020 at 9:25
• Thanks trancelocation Got it. Jan 2, 2020 at 9:12

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$$=(2*0)*e^0+(0^2+1)*e^0=1.$$