Limit Evaluation of the form inf/inf

Question

$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^\alpha} ; \quad \alpha >0$$

Wouldn't it be just zero because $x^\alpha$ will reach $\infty$ much earlier than $\log(x)$?

I tried using L'Hospital's rule but was not able to figure it. Any help will be appreciated.

Thanks

Answer 1

Assuming $\log$ means the natural logarithm, L'Hospital's gives

$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^\alpha }=\lim_{x \rightarrow \infty} \frac{1/x}{\alpha x^{\alpha-1} }=\dfrac{1}{\alpha}\lim_{x \rightarrow \infty} \frac{1}{x^{\alpha} }$$

Answer 2

Without L'Hospital:

Any high-school student has learnt that $\;\lim_{x\to\infty}\dfrac{\ln x}x=0$ (which is proved without the help of the not-so-divine Marquis…).

Now $\ln x=\frac 1\alpha\ln(x^\alpha)$, so, setting $u=x^\alpha$, $\lim_{x\to \infty}u=\infty$, and $$\frac{\ln x}{x^\alpha}=\frac1\alpha\frac{\ln u}u\xrightarrow[u\to \infty]{}\frac1\alpha\cdot 0=0.$$

In the same vein, it is easy to show that for any $\alpha,\beta>0$, one has

$\qquad\qquad\displaystyle\lim_{x\to\infty}\frac{\ln^\beta > x}{x^\alpha}=0,\quad\text{i.e.}\quad \ln^\beta > x=_{\infty}o\bigl(x^\alpha\bigr).$