Is this limit indeterminate or $e^2$ or what? What is the answer to this:

$$
\lim_{x\to ∞} \left({2x+3\over 2x-1}\right)^x
$$

My calculator says this is $ e^2 $ but the only answer I can get to is $ 1^\infty $, which is indeterminate.
 A: $$\lim_{x\to+\infty}\left(\frac{2x+3}{2x-1}\right)^x=\lim_{x\to+\infty}\left(\frac{2x-1+4}{2x-1}\right)^x=\lim_{x\to+\infty}\left(1+\frac{4}{2x-1}\right)^x=$$
$$\lim_{x\to+\infty}\left(\left(1+\frac{1}{\frac{2x-1}{4}}\right)^\frac{2x-1}{4}\right)^\frac{4x}{2x-1}=e^2$$
since:
I. $$\lim_{x\to+\infty}\frac{2x-1}{4}=+\infty$$
and then $$\lim_{x\to+\infty}\left(1+\frac{1}{\frac{2x-1}{4}}\right)^\frac{2x-1}{4}=e$$
II. $$\lim_{x\to+\infty}\frac{4x}{2x-1}=\lim_{x\to+\infty}\frac{4}{2-\frac{1}{x}}=2$$
A: Using the limit definition of the exponential function
$$e^z=\lim_{x\to \infty}\left(1+\frac zx\right)^x$$
we can write
$$\begin{align}
\lim_{x\to \infty}\left(\frac{2x+3}{2x-1}\right)^x&=\lim_{x\to \infty}\left(\frac{1+\frac{3/2}{x}}{1+\frac{-1/2}{x}}\right)^x\\\\
&=\frac{\lim_{x\to \infty}\left(1+\frac {3/2}{x}\right)^x}{\lim_{x\to \infty}\left(1+\frac {-1/2}{x}\right)^x}\\\\\
&=\frac{e^{3/2}}{e^{-1/2}}\\\\
&=e^2
\end{align}$$
A: Consider $$A= \left({2x+3\over 2x-1}\right)^x$$ $$\log(A)=x\log\left({2x+3\over 2x-1}\right)=x\log\left(1+{4\over 2x-1}\right)$$ Now, remembering that, for small $y$, $\log(1+y)\sim y$ $$\log(A)\sim x \times {4\over 2x-1}={4x\over 2x-1}\sim {4x\over 2x}=2$$
A: $$\left(\frac{2x+3}{2x-1}\right)^x=\left(\frac{1+\frac{3}{2x}}{1+\frac{-1}{2x}}\right)^x=\frac{\left[\left(1+\frac{1}{\frac{2x}{3}}\right)^{\frac{2x}{3}}\right]^{\frac{3}{2}}}{\left[\left(1+\frac{1}{-2x}\right)^{-2x}\right]^{\frac{-1}{2}}}$$
Now, noting that$\left(1+\frac{1}{y}\right)^y\to e$ as $\vert y \vert \to \infty$, we can take $y=\frac{2x}{3}, -2x$ in the numerator and denominator respectively to see that this limit tends to $\frac{e^{\frac{3}{2}}}{e^{\frac{-1}{2}}}=e^2$ as required.
A: $$\left(\frac{2x+3}{2x-1}\right)^x=\exp\left(x\ln\left(1+\frac{4}{2x-1}\right)\right)$$
We have $\ln\left(1+\frac{4}{2x-1}\right)=\frac{4}{2x-1}+o_{+\infty}\left(\frac{1}{x}\right)$. 
Then $x\ln\left(1+\frac{4}{2x-1}\right)=\frac{4x}{2x-1}+o(1)\to 2$ as $x\to+\infty$.
Hence $\lim\limits_{x\to +\infty}\left(\frac{2x+3}{2x-1}\right)^x=e^2$
