Limit of logarithmic function using l'Hospital How can I find the following limit:
$$\lim_{x\rightarrow \infty}\frac{\ln(1+\alpha x)}{\ln(\ln(1+\text{e}^{\beta x}))}$$
where $\alpha, \ \beta \in \mathbb{R}^+$.
My first guess was to use l'Hospital:
$$\lim_{x\rightarrow \infty}\frac{\ln(1+\alpha x)}{\ln(\ln(1+\text{e}^{\beta x}))} = \lim_{x\rightarrow \infty}\frac{\ln(1+\text{e}^{\beta x})(1+\text{e}^{\beta x}) \ \alpha}{(1 + \alpha x) \ \text{e}^{\beta x} \  \beta}$$
But what can I do now? Is my approach correct or is there a simpler method?
Edit: Taking the advice from Daniel Fischer, 
$$\lim_{x\rightarrow \infty}\frac{\ln(1+\text{e}^{\beta x})(1+\text{e}^{\beta x}) \ \alpha}{(1 + \alpha x) \ \text{e}^{\beta x} \  \beta} = \lim_{x\rightarrow \infty}\frac{\ln(1+\text{e}^{\beta x}) \ \alpha}{(1 + \alpha x)  \  \beta}  \lim_{x\rightarrow \infty}(1+\text{e}^{-\beta x}) $$ 
Applying L'Hospital a second time on the first fraction, 
$$\lim_{x\rightarrow \infty}\frac{\ln(1+\text{e}^{\beta x}) \ \alpha}{(1 + \alpha x)  \  \beta}  \lim_{x\rightarrow \infty}(1+\text{e}^{-\beta x}) = \lim_{x\rightarrow \infty}\frac{\alpha  \ \beta \  \text{e}^{\beta x}}{(1+\text{e}^{\beta x}) \ \alpha \ \beta}    \cdot 1 = \lim_{x\rightarrow \infty}\frac{\text{e}^{\beta x}}{1+\text{e}^{\beta x} }    \cdot 1  $$
Now let's apply L'Hospital one final time:
$$\lim_{x\rightarrow \infty}\frac{\text{e}^{\beta x}}{1+\text{e}^{\beta x} }    \cdot 1 = \lim_{x\rightarrow \infty}\frac{\text{e}^{\beta x}\ \beta}{\text{e}^{\beta x}\ \beta}    \cdot 1 = 1$$
Is this correct?
 A: without L'Hospital, but with Maclaurin series: 
1) rewrite the numerator as $\log (\alpha x) + \log (1+ \frac{1}{\alpha x}) \sim \log \alpha + \log x + \frac{1}{\alpha x}$
2) rewrite the numerator as $\log (\log e^{\beta x} + \log (1+ \frac{1}{\beta x})) = \log (\beta x + \frac{1}{\beta x}) = \log \beta x(1+\frac{1}{(\beta x)^2}) \sim \log \beta +\log x + \frac{1}{(\beta x )^2}$
Can you handle from here? 
A: The first application of l'Hôpital's theorem gives
$$
\lim_{x\to\infty}
  \frac{\alpha}{1+\alpha x}
  \left(\frac{\beta e^{\beta x}\big/(1+e^{\beta x})}{\log(1+e^{\beta x})}\right)^{-1}
=
\lim_{x\to\infty}
  \frac{\alpha}{\beta}
  \frac{1+e^{\beta x}}{e^{\beta x}}
  \frac{\log(1+e^{\beta x})}{1+\alpha x}
$$
Now,
$$
\lim_{x\to\infty}\frac{1+e^{\beta x}}{e^{\beta x}}=
\lim_{x\to\infty}(e^{-\beta x}+1)=1
$$
so we just need to compute
$$
\lim_{x\to\infty}\frac{\log(1+e^{\beta x})}{1+\alpha x}=
\lim_{x\to\infty}\frac{\beta e^{\beta x}/(1+e^{\beta x})}{\alpha}=
\lim_{x\to\infty}\frac{\beta}{\alpha}\frac{e^{\beta x}}{1+e^{\beta x}}=
\frac{\beta}{\alpha}
$$
You don't need anything new for this limit, because you have just computed the reciprocal.
A: We can proceed as follows $$\begin{aligned}L &= \lim_{x \to \infty}\frac{\log(1 + \alpha x)}{\log(\log(1 + e^{\beta x}))}\\
&= \lim_{x \to \infty}\dfrac{\log\left(\dfrac{1 + \alpha x}{\alpha x}\right) + \log \alpha x}{\log\left(\log\left(\dfrac{1 + e^{\beta x}}{e^{\beta x}}\right) + \beta x\right)}\\
&= \lim_{x \to \infty}\frac{y + \log \alpha x}{\log(z + \beta x)} \text{ where }y = \log((1 + \alpha x)/\alpha x), z = \log((1 + e^{\beta x})/e^{\beta x})\\
&= \lim_{x \to \infty}\frac{y + \log \alpha x}{\log(z + \beta x) - \log \beta x + \log \beta x}\\
&= \lim_{x \to \infty}\frac{y + \log \alpha x}{\dfrac{z}{c} + \log \beta x}\text{ where }c \in (\beta x, \beta x + z)\text { by Mean Value Theorem}\\
&= \lim_{x \to \infty}\frac{y + \log \alpha x}{t + \log \beta x}\text{ where }t = z/c\\
&= \lim_{x \to \infty}\frac{y + \log \alpha + \log x}{t + \log \beta + \log x}\\
&= \lim_{x \to \infty}\dfrac{\dfrac{y}{\log x} + \dfrac{\log \alpha}{\log x} + 1}{\dfrac{t}{\log x} + \dfrac{\log \beta}{\log x} + 1}\\
&= \frac{0 + 0 + 1}{0 + 0 + 1} = 1\end{aligned}$$ Here we have used the fact that $\alpha > 0, \beta > 0, y > 0, z > 0$ and as $x \to \infty$ we have $y \to 0$, $z \to 0$, $c \to \infty$ and $t \to 0$.
