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$$\lim_{n\to\infty} n\bigg[e^{\frac x{\sqrt n}}-\frac x{\sqrt n}-1\bigg] = \frac{x^2}{2}$$

I'm not sure how to solve it. hope somebody could help me!

_ Is there any way to see solutions for limit problem?

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  • $\begingroup$ The result should be $$\frac{x^2}{2}$$ $\endgroup$ Jun 27 '18 at 14:06
  • $\begingroup$ @kingW3 Apologies , im sorry for the mistake of mine.glad it has been rectified $\endgroup$ Jun 27 '18 at 14:09
  • $\begingroup$ @TheIntegrator No problem, I was also somewhat harsh I apologize. $\endgroup$
    – kingW3
    Jun 27 '18 at 14:41
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By the change of variable $t:=x/\sqrt n$ you can write

$$ \lim_{n\to\infty}n\left(e^{x/\sqrt n}-\frac x{\sqrt n}-1\right)= x^2\lim_{t\to0^+}\frac{e^t-t-1}{t^2}$$ and the dependency on $x$ is gone.

Now by L'Hospital, twice, $$\lim_{t\to0^+}\frac{e^t-t-1}{t^2}=\lim_{t\to0^+}\frac{e^t-1}{2t}=\lim_{t\to0^+}\frac{e^t}2.$$


Without L'Hospital, assuming the limit exists,

$$2L-L=2\lim_{2t\to0^+}\frac{e^{2t}-2t-1}{4t^2}-\lim_{t\to0^+}\frac{e^t-t-1}{t^2}=\lim_{t\to0^+}\frac{2e^{2t}-4e^t+2}{4t^2}=\frac12\lim_{t\to0^+}\left(\frac{e^t-1}t\right)^2$$ and the last limit is known to have the value $1^2$.

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  • $\begingroup$ How have you combined limits that depend on $2t$ and $t$ respectively? $\endgroup$
    – adfriedman
    Jun 27 '18 at 17:58
  • $\begingroup$ Just a minor point, the second solution assumes that the limit exists. $\endgroup$ Jun 28 '18 at 6:10
  • $\begingroup$ @ParamanandSingh: that's right. I don't know how to work around this. $\endgroup$
    – user65203
    Jun 28 '18 at 8:17
  • $\begingroup$ Well, for existence I don't know any way simpler than Taylor or L'Hospital's Rule. One can use the binomial expansion on $((1+x/n)^n-1-x)/x^2$ and take limits as $n\to\infty$ followed by $x\to 0$ but the analysis involved is not simpler. $\endgroup$ Jun 28 '18 at 13:46
  • $\begingroup$ @adfriedman: $2t\to0^+$ and $t\to0^+$ are synonyms. $\endgroup$
    – user65203
    Jun 28 '18 at 13:51
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Sice $e^{z/\sqrt{n}}$ is analytic we can consider $$ e^{z/\sqrt{n}} = 1 + \frac{z}{\sqrt{n}} + \sum_{m=2}^\infty \frac{z^m}{\sqrt{n^m}m!} $$ So $$ n\left(e^{z/\sqrt{n}} - 1 - \frac{z}{\sqrt{n}}\right) = \sum_{m=2}^\infty \frac{z^n}{n^{m/2-1}m!} = \frac{z^2}{2} + \sum_{m=3}^\infty \frac{z^m}{n^{m/2-1}m!} $$ Taking limit and changing limit and sum (you can check easily dominate convergence) have the result.

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  • $\begingroup$ You typed $z^n$ instead of $z^m$ $\endgroup$
    – Jakobian
    Jun 27 '18 at 14:44
  • $\begingroup$ There is always another way -_-;; Thank you! $\endgroup$
    – NK Yu
    Jun 28 '18 at 14:06
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For $x\neq 0$ $$\lim_{n\to\infty} n\bigg[e^{\frac x{\sqrt n}}-\frac x{\sqrt n}-1\bigg] \stackrel{y=\frac{x}{\sqrt{n}}}{=} x^2 \lim_{y\to0} \frac{e^y-y-1}{y^2} = x^2 \lim_{y\to0} \frac{e^y-1}{2y} = x^2 \lim_{y\to0} \frac{e^y}{2} = \frac{x^2}{2}$$

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  • $\begingroup$ Thans for your help^^ $\endgroup$
    – NK Yu
    Jun 28 '18 at 14:07

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