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How can I find $$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}};?$$

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

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

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

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 $infty ^0,;$

Hint:

Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$

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

Hint for $xgeq1$:

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 $left(1+frac xnright)^{epsilon n}$
   

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