Finding the limit of $(1+\frac{1}{2n+1})^n$

Question

Using Wolframalpha, the limit is $\sqrt{e}$. Now because we already know $\lim\limits_{n \rightarrow \infty}(1+\frac{1}{n})^n = e$

the term has to be ~$(1+\frac{1}{n})^{\frac{n}{2}}$. My attempt to transform the equation fails though, because im not sure how to get rid of the $-\frac{1}{2}$:

$(1+\frac{1}{2n+1})^n = (1+m)^{\frac{m}{2}-\frac{1}{2}}$

Am I allowed to ignore $-\frac{1}{2}$, because it is a constant, that gets lost for $m \rightarrow \infty$? Is there a different way to get rid of it?

Answer 1

Try $$\lim_{n\to\infty}(\left(1+\dfrac1{2n+1}\right)^n=\left(\lim_{n\to\infty}\left(1+\dfrac1{2n+1}\right)^{2n+1}\right)^{\lim_{n\to\infty}\frac n{2n+1}}$$

Now $\lim_{n\to\infty}\dfrac n{2n+1}=\lim_{n\to\infty}\dfrac1{2+\frac1n}=?$

Alternatively, $$\lim_{m\to0}(1+m)^{\frac{m}{2}-\frac{1}{2}}=\dfrac{\lim_{m\to0}\left((1+m)^m\right)^{1/2}}{\lim_{m\to0}(1+m)^{1/2}}=\dfrac{e^{1/2}}{1^{1/2}}$$

Answer 2

$$\lim\limits_{n\rightarrow\infty}\left(1+\frac{1}{2n+1}\right)^n=\lim\limits_{n\rightarrow\infty}\left(\left(1+\frac{1}{2n+1}\right)^{2n+1}\right)^{\frac{n}{2n+1}}=e^{\frac{1}{2}}$$