<s>[INST] Calculate the limit: $\lim_{x\rightarrow \infty}\frac{\ln x}{x^{a}}$ 
Calculate the limit: $$\lim_{x\rightarrow \infty}\frac{\ln x}{x^{a}}$$

When try calculate limit, we get $\frac{\infty}{\infty}$, so use L'Hôpital again.
$$(\ln x)' = \frac{1}{x}$$
$$x^{a} = e^{\ln x \cdot a} \Rightarrow (e^{\ln x \cdot a})'= e^{\ln x \cdot a} \cdot \frac{1}{x} \cdot a$$
$$\Rightarrow$$
$$\lim_{x\rightarrow\infty}\frac{\frac{1}{x}}{e^{\ln x \cdot a} \cdot \frac{1}{x} \cdot a}= \lim_{x\rightarrow\infty}\frac{x}{e^{\ln x \cdot a} \cdot a \cdot x} = \lim_{x\rightarrow\infty} \frac{1}{e^{\ln x \cdot a} \cdot a} = \frac{1}{\infty} = 0$$
Is correct result and limit?
 [/INST] Why not use the simpler tool called Squeeze Theorem instead? If $a > 0$ then we choose $b$ such that $0 < b < a$ so that $a - b > 0$. Since $x \to \infty$ we can safely assume $x > 1$. And then $x^{b} > 1$ because $b > 0$. We have the inequality $$0 < \log x < x - 1$$ for all $x > 1$ and hence replacing $x$ by $x^{b}$ we get $$0 < b\log x < x^{b} - 1 < x^{b}$$ or $$0 < \log x < \frac{x^{b}}{b}$$ or $$0 < \frac{\log x}{x^{a}} < \frac{1}{bx^{a - b}}$$ By Squeeze theorem we now get $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$
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