Finding the limits of this complicated example $$\lim_{n\to\infty}\left(\frac{2n+3}{2n-1}\right)^{n+1}$$
I have this, but I don't know how to start it.
I tried rationalizing it, or factorizing it but it didn't work.
Is there anyone who can give me any tips on how to do it?
 A: $$\ \lim_{n\to\infty}\bigg(\frac{2n+3}{2n-1}\bigg)^{n+1}=\lim_{n\to\infty}\bigg(\frac{2n-1+4}{2n-1}\bigg)^{n+1}=\lim_{n\to\infty}\bigg(1+\frac{4}{2n-1}\bigg)^{n+1}=$$
$$\ =\lim_{n\to\infty}\bigg[\bigg(1+\frac{4}{2n-1}\bigg)^{2n+2}\bigg]^{\frac{1}{2}}=\lim_{n\to\infty}\bigg[\bigg(1+\frac{4}{2n-1}\bigg)^{2n-1}\cdot\bigg(1+\frac{4}{2n-1}\bigg)^{3}\bigg]^{\frac{1}{2}}=$$
$$=(e^4)^{\frac{1}{2}}\cdot(1)^{\frac{3}{2}}{}=e^2$$
Since $\ \forall\alpha,\beta \in \mathbb R, \lim_{a_n\to\infty}(1+\frac{\alpha}{a_n})^{\beta}=1$
A: Let $m=\frac{2n-1}{4}$.
$$\begin{align}\lim_{n\to\infty}\left(\frac{2n+3}{2n-1}\right)^{n+1}&=\lim_{n\to\infty}\left(1+\frac{4}{2n-1}\right)^{n+1}=\\=\lim_{n\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}&=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}\end{align}$$
, because $n\to\infty\iff m\to\infty$.  
$$\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}=\lim_{m\to\infty} \left(\left(1+\frac{1}{m}\right)^m\right)^2\left(1+\frac{1}{m}\right)^{1.5}=e\cdot e\cdot \lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{1.5}=e^2$$
