Limits of the form $\frac{\exp(x)}{x^\alpha} \quad \frac{\ln(x)}{x^\alpha} $

Question

Let $\alpha >0$ proof:

$$ \lim\limits_{x \to \infty} \frac{\exp(x)}{x^\alpha} = \infty \quad (1)$$ and $$\lim\limits_{x \to \infty} \frac{\ln(x)}{x^\alpha} = 0 \quad (2)$$ and $$\lim\limits_{x \to 0^+} \ln(x) \cdot x^\alpha = 0 \quad (3)$$ which is applying the rule of L'Hôpital straightforward.

I saw this exercise on a sheet for an introductionary analysis course and they do not allow the student to even apply continuity arguments. I wonder how to proof all of them under these restrictions:

For (1) i thought of applying taylor series to get an lower bound which is unbounded. I think i can conculde (2) from (1). I do not have a clue how to solve (3) One could substitute $x \to 0^+ $ with $\frac{1}{n} \to 0^+ $ for $n \to \infty$ but i wonder if this is a continuity argument. Further i am looking for different or more elegant approaches to this.

Answer 1

The only necessary knowledge is located in :

$$\lim_{x\to\infty}\frac{\ln(x)}{x}=0\qquad(\star)$$

Indeed, we have for every $x>0$ :

$$\ln(\frac{\exp(x)}{x^\alpha})=x-\alpha\ln(x)=x\left(1-\alpha\frac{\ln(x)}{x}\right)\underset{x\to\infty}{\longrightarrow}\infty$$

and thus :

$$\lim_{x\to\infty}\frac{\exp(x)}{x^\alpha}=\infty$$ which is the conclusion for (1). For (2), we can use the following relation :

$$\frac{\ln(x)}{x^\alpha}=\frac{1}{\alpha}\,\frac{\ln(x^\alpha)}{x^\alpha}$$

and composition of limits gives us the conclusion.

For (3), we see that for all $x>0$ :

$$x^\alpha\ln(x)=-\frac{1}{\alpha}\frac{\ln(\frac{1}{x^\alpha})}{\frac{1}{x^\alpha}}$$

and again, composition of limits leads to the result.

Hoping this will help.

Note that we have to know first that $(\star)$ holds. This can be proved as follows :

Consider $\phi(x)=\frac{\ln(x)}{x}$, for $x>0$. It's easy to see that $\phi$ is decreasing for $x\ge e$. Since $\phi$ is positive on the interval $[e,+\infty[$, it has a finite limite $L$ at $+\infty$. For all $x>0$ :

$$\frac{\ln(x)}{x}=\frac{2}{\sqrt x}\,\frac{\ln(\sqrt x)}{\sqrt x}$$

and taking the limit (as $x\to\infty$) we get :

$$L=0\times L$$