Prove $\lim_{n\to\infty}\left(1+\frac{1}{2n+50}\right)^n=\sqrt{e}$ I would like to argue that
$$
\lim_{n\to\infty}\left(1+\frac{1}{2n+50}\right)^n=\sqrt{e}.
$$

My idea is that we can neglect the constant value $50$ in the denominator and to substitute $m=2n \Leftrightarrow n=m/2$, i.e.,
$$
\lim_{n\to\infty}\left(1+\frac{1}{2n+50}\right)^n = \lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^n=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{\frac{m}{2}}=\lim_{m\to\infty}\left[\left(1+\frac{1}{m}\right)^m\right]^\frac{1}{2}
$$
Now, I think that I can put the limit inside the brackets, because the function $x\mapsto (1+\frac{1}{x})^{m/2}$ is continuous on $(0,\infty)$, thus
$$
\lim_{m\to\infty}\left[\left(1+\frac{1}{m}\right)^m\right]^\frac{1}{2}=\left[\underbrace{\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^m}_{=e}\right]^\frac{1}{2}=\sqrt{e}.
$$

You have the right idea, but I would say that the way you wrote it, is not a formal argument. It's true that $50$ is "irrelevant", but one needs to be careful about those things.
Here is a more formal argument, using your ideas. You have
$$\tag1
\left(1+\frac{1}{2n+50}\right)^n =
\Bigg[\left(1+\frac{1}{2n+50}\right)^{2n +50}\Bigg]^{1/2}
\left(1+\frac{1}{2n+50}\right)^{-25}.
$$
We focus on the first factor. Because the function $t\longmapsto t^{1/2}$ is continuous (and everything is positive here) we can exchange the limit with the square root. Then
$$\tag2
\lim_{n\to\infty}\Bigg[\left(1+\frac{1}{2n+50}\right)^{2n +50}\Bigg]^{1/2}
=\Bigg[\lim_{n\to\infty}\left(1+\frac{1}{2n+50}\right)^{2n +50}\Bigg]^{1/2}=e^{1/2}.
$$
The second factor in $(1)$ converges to $1$, since the power is fixed. As the limit of the product is the product of the limit (when the limits exist),
$$
\lim_{n\to\infty}\left(1+\frac{1}{2n+50}\right)^{n }=e^{1/2}.
$$
