I have a problem when I go to calculate $\lim_{x\to\infty}\left( \frac {2x+a}{2x+a-1}\right)^{x}.$ The limit:
$\lim_{x\rightarrow\infty}
 \left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$
I make this:
$\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$=${{\rm e}^{{\it x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$
Then:
${{\rm e}^{{\it \lim_{x\rightarrow\infty}
 x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$ = $0\cdot\infty$
Note: I cannot use L'Hopital

One may recall that, as $u \to 0$, by using the Taylor series expansion,
$$
\log(1+u)=u+O(u^2)
$$ one gets, as $x \to \infty$, 
$$
\begin{align}
\log\left(\frac{2x+a}{2x+a-1}\right)&=\log\left(1+\frac1{2x+a-1}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{(2x+a-1)^2}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{x^2}\right)
\end{align}
$$ thus, as $x \to \infty$,
$$
x\log\left(\frac{2x+a}{2x+a-1}\right)=\frac{x}{2x+a-1}+O\left(\frac1{x}\right) \to \color{blue}{\frac12}
$$ and, as $x \to \infty$,
$$
\left(\frac{2x+a}{2x+a-1}\right)^x=e^{x\log\left(\frac{2x+a}{2x+a-1}\right)} \to \color{blue}{\sqrt{e}.}
$$
