<s>[INST] 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
 [/INST] $(\frac{2x+a}{2x+a-1})^x=(\frac{2x+a+1-1}{2x+a-1})^x=(1+\frac{1}{\alpha})^{\frac{\alpha}{2}+\frac{1-a}{2}}\longrightarrow e^{\frac{1}{2}}$. Where $\alpha=2x+a-1$ and $\alpha\rightarrow \infty,$ when $x\rightarrow\infty$
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