How to prove $\lim\limits_{x\to 0}x^m (\ln x)^n = 0$ $$\lim\limits_{x\to 0^+}x^m (\ln x)^n = 0\quad \,\text{for} \quad m,n \in \mathbb N$$
Question: How can I prove this?
Is there a better way than saying, well, if the factor $\lim\limits_{x\to 0}x = 0$ the whole equation is $0$?
Note: I have found a similiar post but however I need to solve the task without integrals.
I appreciate every hint.
 A: Changing variable $y=-\ln x$ the limit becomes
$$(-1)^n\lim_{y\to+\infty}e^{-my}y^n.$$
Since for positive $z$
$$e^z=\sum_{k=0}^\infty \frac{z^k}{k!}\geq \frac{z^{n+1}}{(n+1)!}$$
putting $z=my$ we obtain
$$ e^{-my}y^n=\frac{y^n}{e^{my}}\leq\frac{y^n}{\frac{1}{(n+1)!}(my)^{n+1}}=\frac{(n+1)!}{m^{n+1}}\frac1y$$
and this goes to zero as $y\to+\infty$. Since $e^{-my}y^n$ is positive, by the squeeze theorem the original limit is zero.
A: The issue here is that $\lim_{x\rightarrow 0} x = 0$, but isn't it possible that $\ln(x)$ approaches $-\infty$ faster than $x$ goes to $0$? For instance, consider $e^x/x$. Certainly this function approaches $\infty$ as $x$ goes to infinity, but $1/x$ approaches $0$ as $x$ goes to infinity. So $\lim_{x\rightarrow 0}=0$ is not a sufficient reason for the larger limit to be $0$.
The real question to consider: what tools do you have for solving limits? Do you have the $\epsilon$-$\delta$ definition? L'Hopital's Rule (as asked and answered above)? Basically, what are the rules of the game for this question? (I ask, because there are probably multiple ways to solve that problem.)
A: Let $x=e^{-t}$. Then
$$L=\lim_{t\to\infty}{e^{-mt}t^n}=\left(\lim_{t\to\infty}{e^{-mt/n}t}\right)^n=\left(\frac nm\lim_{t\to\infty}{e^{-t}t}\right)^n.$$
Now by induction,
$$t\ge3\implies e^{-t-1}(t+1)=\frac{t+1}{et}e^{-t}t<\frac12e^{-t}t$$ and the limit is $0$.
A: Write $$x^m\ln^nx=e^{m\ln x+\ln\ln^n x}:=e^{\varphi(x)},\quad \:\varphi(x)=m\ln x\left(1+{\ln \ln^n x\over m\ln x}\right).$$
We have that
$$\forall x>0\:\:\:\forall \varepsilon>0\:\:\exists N_{\varepsilon,x}>0\:\:\forall n\ge N_{\varepsilon,x}\implies \left|m\ln x-\varphi (x)\right|\le \varepsilon\:|m\ln x|.$$ $$$$
So that $\:\varphi(x)\underset{\:0^+}{\sim}\:m\ln x\:$, since $\:\lim_{x\to 0^+}\Large(\normalsize\varphi(x)/(m\ln x)-1\Large)\normalsize\:=0.$ $$$$
By now it's a piece of cake to prove that
$$ \forall (m,n)\in \mathbf N^2\:,\:\: e^{\varphi(x)}\underset {\:0^+}\sim\:e^{\ln x^m}=\:x^m\overset{\:x\to 0^+}\longrightarrow \:0.$$
A: The ideal solution here is highly dependent on the mathematical toolbox available to you. I present a solution here which is not too technical (easing off on calculus where possible), at the cost of a little rigor:
Claim:
$$\lim\limits_{x\to 0^+}x^m (\ln x)^n = 0, \,\text{for} \quad m,n \in \mathbb N$$
Remark: Taking $n^{th}$ roots, we note that, if we can show $x^\alpha\ln x \to0$ for all (rational) $\alpha>0$, then we have our result.
Observation: $$\lim_{x\to0^+}x^\alpha\ln x=\lim_{1/x\to0}(1/x)^\alpha\ln(1/x)=\lim_{x\to\infty}\left[-\frac{\ln x}{x^\alpha}\right]$$
Now, admittedly this is not totally trivial - it relies on the continuity of the functions as hand, as well as being able to convince yourself that we should be allowed to shift limits at $0$ to limits at infinity this way. However, once you can come to terms with this being true (perhaps luckily, it is), we're not too far from a solution.
Lemma:
$$\ln x \le x-1 < x \text{ for } x>0$$
There are many proofs of this, though in honesty, most use calculus or differentiation at some point or another. Still, this is a fairly uncontroversial inequality, so we use it to establish:
Corollary:
$$\ln\left(x^{1/m}\right)<x^{1/m}\implies \ln x < mx^{1/m} \text{ for } m>0$$
Bringing it all together:
From this, we can see that $\frac{\ln x}{x^\alpha}<mx^{1/m-\alpha}$ for any positive $m$.
Thus for sufficiently large $m$, this ratio is bounded above by a multiple of $x^{-r}$ for some positive $r$, which tends to 0. 
Given that the ratio is nonnegative for $x>1$, we see that $\frac{\ln x}{x^\alpha}\to 0$ as $x\to\infty$.
Retracing our steps, we can see that our claimed inequality must be true.
A: We want to show $x^m(\ln x)^n \to 0,$ which is the same as showing $x^m|\ln x|^n \to 0.$ Apply $\ln $ to see this is the same as showing
$$\tag 1 m\ln x + n \ln (|\ln x|) \to -\infty.$$
Now for $u\ge 4,$ $\ln u \le u^{1/2}.^*$ And for small $x>0,$ $|\ln x| \ge 4.$ For such $x$ the left side of $(1)$ is bounded above by
$$m\ln x + n |\ln x|^{1/2} =  -m|\ln x| + n |\ln x|^{1/2} = |\ln x|^{1/2}(-m|\ln x|^{1/2} + n).$$ 
As $x\to 0^+,$ the last expression is looking like $\infty\cdot (-\infty)=-\infty.$  This proves $(1)$ and we're done.
$^*$To prove this, note that it's true at $4,$ and then compare derivatives.
