# 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 $x \to \infty$, $$\left(\frac{2x+a}{2x+a-1}\right)^x=e^{x\log\left(\frac{2x+a}{2x+a-1}\right)} \to \color{blue}{\sqrt{e}.}$$
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 $x\rightarrow\infty$