# How can I find $\lim_{n\rightarrow\infty}(1+\frac{x}n)^{\sqrt{n}}$?

How can I find $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}}\;?$$

I know $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{n} = \exp (x)$$ but I don't know how can I put the definition in this particular limit.

I know then, that $$\lim_{n\rightarrow\infty}\big(1+\frac{x}n\big)=1$$, but I don't think this is right to consider.

$$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} = \lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}}$$ From $$\lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]=e \quad \text{and} \quad \lim_{n\rightarrow\infty}{{\frac{x}{n}}{\sqrt{n}}}=0,$$ **, we get $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} = e^0=1$$

EDIT
I add the note bellow as my calculation was considered insufficiently justified

**and because the terms are positive, and we don't have an indeterminate case $$0^0$$ or $$1^{\infty}$$ or $$\infty ^0,\;$$

• One needs an argument for "if $a_n\to a$ and $b_n\to b$ ($b_n,b\geqslant0, a_n,a>0$) then $a_n^{b_n}\to a^b$". – user587192 Nov 20 '18 at 15:12
• Or, if you prefer, put $t={n \over x}.$ The mentioned limit becomes $\lim_{t \to \infty} (1+{1\over t})^{t}.$ – user376343 Nov 20 '18 at 15:18
• What do you use for the "we get..." step in the last line? – user587192 Nov 20 '18 at 15:19
• @user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $\lim_n a_n^{b_n}$ into $(\lim_n a_n)^{(\lim_n b_n)}$ but this isn't necessarily justified. – Jam Nov 20 '18 at 15:26
• From $$\lim_{n\to\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]=e \quad \text{and} \quad \lim_{n\to\infty}{{\frac{x}{n}}{\sqrt{n}}}=0,$$ we get $$\lim_{n\to\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} =\lim_{n\to\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}} \color{red}{ =\left[\lim_{n\to\infty}\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{\lim_{n\to\infty}{\frac{x}{n}}{\sqrt{n}}} } \color{blue}{= e^0}$$ You have the blue one, but you need the red part, which is not explicitly justified. – user587192 Nov 20 '18 at 15:44

Let $$y=(1+\frac{x}{n})^{\sqrt{n}}$$. Note that $$\ln y=\sqrt{n}\ln(1+\frac{x}{n})=\frac{\ln(1+\frac{x}{n})}{1/\sqrt{n}}.$$ As $$n\to\infty$$, this is a limit of the form $$0/0$$. Thus, we may apply L'Hopital's rule to obtain that $$\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{-\frac{x}{n^{2}}/(1+\frac{x}{n})}{-\frac{1}{2}n^{-3/2}}=\lim_{n\to\infty}\frac{2x}{\sqrt{n}+\frac{x}{\sqrt{n}}}=0$$ Thus, $$\lim_{n\to\infty}y=\lim_{n\to\infty}e^{\ln(y)}=e^{\left(\lim_{n\to\infty}\ln(y)\right)}=e^{0}=1.$$

Hint:

Compute the limit of the log: $$\;\sqrt n\log\Bigl(1+ \dfrac xn\Bigr)$$ and use equivalence: $$\log(1+u)\sim_0 u.$$

• I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $\log(1+u)$ with $u$. Since $\log(1+u)\sim u$, $\lim_{u\to\infty}\frac{\log(1+u)}{u}=1$ so $\lim_{n\to\infty}\frac{\log(1+x/n)/\left(\frac{x}{n}\right)}{\sqrt{n}/{\left(\frac{x}{n}\right)}}=\frac{\lim(\ldots)}{\lim_{n\to\infty}\sqrt{n}{\left(\frac{n}{x}\right)}}$. – Jam Nov 20 '18 at 15:32

Hint for $$x\geq1$$:

1. The expression and therefore the limit is relatively easily seen to be greater than or equal to $$1$$
2. The expression is for any $$\epsilon>0$$ eventually smaller than $$\left(1+\frac xn\right)^{\epsilon n}$$

By L'Hopital, $$\lim_{y\to\infty}\sqrt{y}\log(1+\frac{x}{y}) =\lim_{t\to 0}\frac{\log(1+tx)}{\sqrt{t}} =\lim_{t\to 0}\dfrac{\frac{x}{1+tx}}{\frac{1}{2\sqrt{t}}}=0.$$ Now one can use the continuity of the exponential function: $$\lim_{n\to\infty}\exp\left[\sqrt{n}\log(1+\frac{x}{n})\right] =\exp\left[\lim_{n\to\infty}\sqrt{n}\log(1+\frac{x}{n})\right]=1.$$