How to prove $\lim\limits_{x\to 0}x^m (\ln x)^n = 0$

Solution 1:

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.

Solution 2:

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 $0$.

Solution 3:

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.)

Solution 4:

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

$$ \forall (m,n)\in \mathbf N^2\:,\:\: e^{\varphi(x)}\underset {\:0^+}\sim\:e^{\ln x^m}=\:x^m\overset{\:x\to 0^+}\longrightarrow \:0.$$