# I got stuck at limit problem

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

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

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

• The result should be $$\frac{x^2}{2}$$ Jun 27 '18 at 14:06
• @kingW3 Apologies , im sorry for the mistake of mine.glad it has been rectified Jun 27 '18 at 14:09
• @TheIntegrator No problem, I was also somewhat harsh I apologize. Jun 27 '18 at 14:41

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 $1^2$.

• How have you combined limits that depend on $2t$ and $t$ respectively? Jun 27 '18 at 17:58
• Just a minor point, the second solution assumes that the limit exists. Jun 28 '18 at 6:10
• @ParamanandSingh: that's right. I don't know how to work around this.
– user65203
Jun 28 '18 at 8:17
• 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. Jun 28 '18 at 13:46
• @adfriedman: $2t\to0^+$ and $t\to0^+$ are synonyms.
– user65203
Jun 28 '18 at 13:51

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.

• You typed $z^n$ instead of $z^m$ Jun 27 '18 at 14:44
• There is always another way -_-;; Thank you! Jun 28 '18 at 14:06

For $x\neq 0$ $$\lim_{n\to\infty} n\bigg[e^{\frac x{\sqrt n}}-\frac x{\sqrt n}-1\bigg] \stackrel{y=\frac{x}{\sqrt{n}}}{=} x^2 \lim_{y\to0} \frac{e^y-y-1}{y^2} = x^2 \lim_{y\to0} \frac{e^y-1}{2y} = x^2 \lim_{y\to0} \frac{e^y}{2} = \frac{x^2}{2}$$

• Thans for your help^^ Jun 28 '18 at 14:07