Proof that $\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$ without L'Hospital I proved that $$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$$
using L'Hospital's rule. But is there a way to prove it without L'Hospital's rule? I tried splitting it as
$$\lim_{n\to\infty}n^{-n}(n^2+x^2)^{\frac{n}{2}},$$
but that didn't work because $\lim_{n\to\infty}(n^2+x^2)^{\frac{n}{2}}$ diverges.
 A: This has the form $\displaystyle\lim_{n\to\infty} (1+1/n)^{n}=e$. 
$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}{\color{red} {\frac{n}{x^2}\cdot\frac{x^2}{n}} }}=\lim_{n\to\infty}\left(\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{x^2}}\right)^{{\color{red} {\frac{x^2}{2n}} }}=e^{\displaystyle\lim_{n\to\infty} \frac{x^2}{2n}} = e^0 = 1$$
Note Since $n\to\infty$ then $1/n^2$ has the same behaivor that $1/(n^2/x^2) = x^2/n^2$.
A: METHODOLOGY $1$:  Direct Application of Bernoulli's Inequality
Note that for $n>|x|$
$$1\le \left(1+\frac{x^2}{n^2}\right)^{n/2}\le \frac1{\left(1-\frac{x^2}{n^2}\right)^{n/2}}\le \frac1{1-\frac{x^2}{2n}}$$
where we used Bernoulli's inequality to arrive at the last inequality.
Now apply the squeeze theorem to find 
$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$


METHODOLOGY $1$:  Using Estimates of the Logarithm Function
Note that we may write
$$\left(1+\frac{x^2}{n^2}\right)^{n/2}=e^{(n/2)\log\left(1+\frac{x^2}{n^2}\right)}\tag 1$$
In This Answer, I used elementary, pre-calculus tools to obtain the inequalities 
$$\frac{x}{1+x}\le \log(1+x)\le x \tag2$$
Using $(2)$ in $(1)$ reveals
$$e^{nx^2/(2n^2+2x^2)}\le e^{(n/2)\log\left(1+\frac{x^2}{n^2}\right)}\le e^{x^2/2n}$$ 
whence application of the squeeze theorem yields the coveted result
$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$
as expected!
A: Consider the following for large n and finite x:
$$e^{\frac{x^2}{n^2}} \approx 1+\frac{x^2}{n^2}$$ 
Therefore, rewrite the limit as:
$$\lim_{n \to \infty} {\left(e^{\frac{x^2}{n^2}}\right)}^{\frac{n}{2}}$$
$$=\lim_{n \to \infty} e^{\frac{x^2}{2n}}$$
$$=1$$
A: I have an algebraic solution. Let's our limit be $L$:
$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}$$
Now, we make two changes of variables: $$t = \frac{n}{2} $$ and
$$y=\frac{x^2}{4}$$
Then we have:
$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=\lim_{t\rightarrow\infty}\left(1+\frac{x^2}{2t^2}\right)^t=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t$$
Then, rewrite our limit as:
$$L=\lim_{t\rightarrow\infty}\left(1+\frac{y}{t^2}\right)^t=\lim_{t\rightarrow\infty}e^{t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}}=e^{L_1}$$
Where $L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}$
Now, we make another change of variables:
$$r=1/t^2$$
$$L_1=\lim_{t\rightarrow\infty} t \ln{\left(1+\frac{y}{t^2}\right)}=\lim_{t\rightarrow0} \frac{\ln{\left(1+ry\right)}}{\sqrt{r}}=\lim_{r\rightarrow0} \frac{\ln{\left(1+ry\right)}}{yr} \frac{yr}{\sqrt{r}}=\lim_{r\rightarrow0} y\sqrt{r}=0$$
Finally:
$$L=\lim_{n\rightarrow\infty}\left(1+\frac{x^2}{n^2}\right)^\frac{n}{2}=e^{L_1}=e^0=1$$
A: For the upper bound using Bernoulli inequality note that it applies for exponents $t: t \leq 0 \cup t \geq 1$, so for $\frac{n}{2} < 0$:
$$
\bigg(1+\frac{x^2}{n^2} \bigg)^\frac{n}{2}= \frac{1}{\bigg(1+\frac{x^2}{n^2} \bigg)^{-\frac{n}{2}}} \leq \frac{1}{1- \frac{x^2}{2n}} \to 1
$$
And the limit follows to squeeze lemma
A: The lemma of Thomas Andrews can be used here:

Lemma: If $n(a_n-1)\to 0$ then $a_n^n\to 1$.

Now use this with $$a_n=\sqrt{1+\frac{x^2}{n^2}}$$

Perhaps you are trying to deal with the limit of $(1+ix/n)^n$ and show that it equals $\cos x+i\sin x$. That can also be easily handled by the lemma in question without first dealing with $|(1+ix/n)^n|$. Just apply the lemma to $$a_n=\dfrac{1 +\dfrac{ix} {n}} {\cos\dfrac{x} {n} +i\sin\dfrac{x} {n}} $$
