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?
 A: You are assuming $a > 0$, but yes, it is correct. 
Only weird thing is how you calculate the derivative of $x^a$; usually this is part of the "fundamental" derivatives that one learns, and should immediately be recognized as equal to $ax^{a-1}$. Your calculations yield the correct answer but it's a little weird ;-)
A: Actually, even more is true. For $p>0$ and $q>0$, we have that
$$
\lim_{x\to\infty}\frac{\log^px}{x^q}
 =\lim_{x\to\infty}\Bigl(\frac{\log x}{x^{q/p}}\Bigr)^p
 =\Bigl(\lim_{x\to\infty}\frac{\log x}{x^{q/p}}\Bigr)^p
 =\Bigl(\lim_{x\to\infty}\frac1{(q/p)x^{q/p}}\Bigr)^p=0
$$
using continuity and l'Hôpital's rule.
A: 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 $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$
A: Because I'm not a fan of using L'Hospital whenever one has a limit of the Form $\frac{\infty}{\infty}$, here is a way that makes use of the fact that $\ln$ is the inverse function of $\exp$:
We first show the following:

Let $P:\mathbb R\rightarrow\mathbb R$ be a polynomial, then $$\lim\limits_{x\to\infty} \frac{P(x)}{\exp(x)}=\lim\limits_{x\to -\infty} P(x)\exp(x)$$ where $\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto \sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$.

Let $n$ be the degree of the polynomial $P$. We can find $c>0$ and $R>0$ with $$\left|P(x)\right|\leq c|x|^n~\text{for all}~x\geq R.$$ Further we have for $x\geq 0$:
$$\exp(x)=\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}\stackrel{x\geq 0}\geq\frac{x^{n+1}}{(n+1)!}$$
and therefor we have
$$\left|\frac{P(x)}{\exp(x)}\right|\leq \frac{cx^n}{\frac{1}{(n+1)!}x^{n+1}}=\frac{c\cdot(n+1)!}{x}$$
for all $x\geq R$. Using the squeeze theorem we now get:
$$\lim\limits_{x\to\infty}\frac{P(x)}{\exp(x)}=0~\text{and}~\lim\limits_{x\to -\infty} P(x)\exp(x)=\lim\limits_{w\to\infty} \frac{P(-w)}{\exp(w)}=0.$$
We will use this to prove the following:

Let $\alpha\in\mathbb R_+^*$. Then $$\lim\limits_{x\to\infty} \frac{\ln(x)}{x^{\alpha}}=0.$$

Let $(x_n)_{n\in\mathbb N}$ be a sequence in $\mathbb R_+^*$ with $\lim\limits_{n\to\infty} x_n=\infty$. We then also have for $y_n:=\alpha\ln(x_n)$ that $\lim\limits_{n\to\infty} y_n=\infty$. 
As $$x_n^{\alpha}=\exp(y_n)$$ we now get:
$$\lim\limits_{n\to\infty} \frac{\ln(x_n)}{x_n^{\alpha}}=\lim\limits_{n\to\infty} \frac{1}{\alpha}y_n\exp(-y_n)=0.$$
