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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?

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You obtain $\frac{\infty}{\infty}$ after L'Hopital the first time so it doesn't hurt to try L'Hopital again and see where that leads. (Also, I think it should be $(1+\alpha x)e^{\beta x} \beta$ in the denominator which is obtained by LH.) –  Gamma Function Jun 10 at 11:53
2  
You should have $1+ \alpha x$ in the denominator, not $1+\alpha$. Write $\frac{1+e^{\beta x}}{e^{\beta x}} = 1 + e^{-\beta x}$. Since that tends to $1$ as $x\to\infty$, you can remove that to simplify. –  Daniel Fischer Jun 10 at 11:54
    
Thanks for your comments! I edited my question. Can you verify if it's correct or did I make a mistake somewhere? –  IronMan12 Jun 10 at 13:20

3 Answers 3

up vote 1 down vote accepted

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.

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How come that your final result is $\beta / \alpha$ while Paramanand Singh has $1$ as final result? Where did the factor $\alpha/ \beta$ from the first equation go to? I'm a bit confused now. –  IronMan12 Jun 11 at 20:07
    
@IronMan12 I just studied separately the three factors in the final limit of the first display. The first factor is $\alpha/\beta$, the second has limit $1$, the third has limit $\beta/\alpha$. So the product is $1$. –  egreg Jun 11 at 20:09
    
Ah, wonderful. That makes sense. Thank you so much for your fast answer! –  IronMan12 Jun 11 at 20:11

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?

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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 $t \to 0$.

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