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.
 A: $$\lim_{n\rightarrow\infty}\left(1+\frac{x}n\right)^{\sqrt{n}}  = \lim_{n\rightarrow\infty}\left[\left(1+\frac{x}n\right)^{{\frac{n}{x}}}\right]^{{\frac{x}{n}}{\sqrt{n}}}$$ 
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 $\infty ^0,\;$
A: Hint:
Compute the limit of the log: $\;\sqrt n\log\Bigl(1+ \dfrac xn\Bigr)$ and use equivalence:
$$\log(1+u)\sim_0 u.$$
A: 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, 
$$
\lim_{n\to\infty}y=\lim_{n\to\infty}e^{\ln(y)}=e^{\left(\lim_{n\to\infty}\ln(y)\right)}=e^{0}=1.
$$
A: Hint for $x\geq1$:


*

*The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$

*The expression is for any $\epsilon>0$ eventually smaller than $\left(1+\frac xn\right)^{\epsilon n}$
A: 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: 
$$
\lim_{n\to\infty}\exp\left[\sqrt{n}\log(1+\frac{x}{n})\right]
=\exp\left[\lim_{n\to\infty}\sqrt{n}\log(1+\frac{x}{n})\right]=1.
$$
