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$$\lim\limits_{x\to 0^+}x^m (\ln x)^n = 0\quad \,\text{for} \quad m,n \in \mathbb N$$

Question: How can I proof 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.

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    Are you familiar with L'Hopital's rule? – 2017-01-02
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    no, I am not familiar with L'Hospital's rule – 2017-01-02
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    Technically this should be $\lim_{x\rightarrow 0^{+}}x^m(lnx)^n = 0$ as $lnx$ is not defined for $x < 0$. To be honest, I'm not entirely convinced that the statement of the problem itself is true... – 2017-01-02
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    @jublikon It helped if you showed your work on the problem, and pointed out where you got stuck. `I am not familiar with L'Hospital's rule` What is the method you are familiar with to prove the case $m=n=1\,$? – 2017-01-02

6 Answers 6

1

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.

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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.

2

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.)

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    yes, I know the $\epsilon - \delta$ definition – 2017-01-02
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    Okay then with the $\epsilon$-$\delta$ definition of a limit, you could try letting $\epsilon>0$ and show that there exists some $\delta>0$ such that $0 – 2017-01-02
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    Can we do this by l hospital... I don't e-€ definition – 2018-10-02
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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 $0$.

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    Hi - are you doing induction on integer $t$ here? – 2017-01-02
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    @πr8: yep, $t\to t+1$. – 2017-01-02
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    Yep, just checking - I'm not sure this suffices for a continuous limit though? The idea is otherwise sound :) – 2017-01-02
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    @πr8: for the base hypothesis, you can take $3\le t<4\implies e^{-t}t<1$, that's enough. As the function is smooth, proving for integers or for reals is virtually identical. – 2017-01-02
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    Gotcha: $e^t>e^3>4>t$ and the rest follows from that. Thanks for the clarification and +1 – 2017-01-02
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    simple, lucid, beautiful! – 2017-11-09
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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

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

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    liked the way! so simple. – 2017-11-09
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    It's my pleasure to help you fellow $\text{mathematician}$. – 2017-11-09
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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:

$$\ln\left(x^{1/m}\right)0$$

Bringing it all together:

From this, we can see that $\frac{\ln x}{x^\alpha}

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.