Proof that $\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}}=1$ without L'Hospital

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 $1/(n^2/x^2) = x^2/n^2$.

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

$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$

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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!

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: $$\lim_{n \to \infty} {\left(e^{\frac{x^2}{n^2}}\right)}^{\frac{n}{2}}$$ $$=\lim_{n \to \infty} e^{\frac{x^2}{2n}}$$ $$=1$$