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 $x \to \infty$, $$ \left(\frac{2x+a}{2x+a-1}\right)^x=e^{x\log\left(\frac{2x+a}{2x+a-1}\right)} \to \color{blue}{\sqrt{e}.} $$
Hint: $$ \frac{2x+a}{2x+a-1} = 1 + \frac{1}{2x+a-1} $$
Does that look a little similar to the following now? $$ \lim_{n->\infty} \left(1+\frac{1}{n}\right)^n $$
$(\frac{2x+a}{2x+a-1})^x=(\frac{2x+a+1-1}{2x+a-1})^x=(1+\frac{1}{\alpha})^{\frac{\alpha}{2}+\frac{1-a}{2}}\longrightarrow e^{\frac{1}{2}}$. Where $\alpha=2x+a-1$ and $\alpha\rightarrow \infty,$ when $x\rightarrow\infty$
