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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

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4 Answers 4

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

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Option:

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

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

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

$1+O(1/n)$;

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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,

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

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  • $\begingroup$ Thanks Trancelocation please explain me how i prove $e^{x}<1+xe^{x}$ for $x>0$ $\endgroup$
    – jacky
    Jan 1, 2020 at 9:17
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    $\begingroup$ @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 $$xe^0 < e^x-1 < xe^x$$ $$\Leftrightarrow 1+x < e^x < 1+xe^x$$ $\endgroup$ Jan 1, 2020 at 9:25
  • $\begingroup$ Thanks trancelocation Got it. $\endgroup$
    – jacky
    Jan 2, 2020 at 9:12
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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$=(2*0)*e^0+(0^2+1)*e^0=1.$

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