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

Question

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.

Answer 1

$$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}} = \lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}}$$ 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 $\infty ^0,\;$

Answer 2

Hint:

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

Answer 3

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 $\left(1+\frac xn\right)^{\epsilon n}$

Answer 4

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, $$ \lim_{n\to\infty}y=\lim_{n\to\infty}e^{\ln(y)}=e^{\left(\lim_{n\to\infty}\ln(y)\right)}=e^{0}=1. $$

Answer 5

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: $$ \lim_{n\to\infty}\exp\left[\sqrt{n}\log(1+\frac{x}{n})\right] =\exp\left[\lim_{n\to\infty}\sqrt{n}\log(1+\frac{x}{n})\right]=1. $$