How to show that $\lim_{x \to 0} x^p (\ln x)^r = 0$

Question

I want to show that $\lim_{x \to 0} x^p (\ln x)^r = 0$ if $p > 1$ and $r \in \mathbb{N}$.

To show this, I wanted to use that $\lim_{x \to 0} x \ln x = 0$, and in fact if $p \geq r$ we can write that $$\lim_{x \to 0} x^p (\ln x)^r = \lim_{x \to 0} x^{p-r}((x \ln x)^r)$$ and both terms of the product goes to 0 as $x \to 0$. Thus this part is okay.

I have trouble if $p < r$. I tried to do the same, but I have no result so far. Maybe by considering $(x \ln x)^p (\ln x)^{r-p}$ and then setting $y = \ln x$. Thus we obtain $$ \begin{align*} \lim_{x \to 0} (x \ln x)^p (\ln x)^{r-p} &= \lim_{e^y \to 0} (e^y \cdot y)^p \cdot y^{r-p} \\ &= \lim_{y \to - \infty} (e^y \cdot y)^p \cdot y^{r-p} \\ &= \lim_{y \to - \infty} e^{py} y^r\\ &= 0 \end{align*} $$ because the exponential goes "faster" to $0$ than a polynom diverges to infinity? Does it seem correct to you?

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Edit:

I will assume that we know that for $r \in \mathbb{N}$, $$\lim_{y \to \infty} \frac{y^r}{e^y} = 0.$$ Step 1: I will prove that for $p > 1$ and $r \in \mathbb{N}$ (for my exercise), $$\lim_{x \to \infty} \frac{\ln^r x}{x^p} = 0.$$ Proof: Let $y = \ln x$. Then $x = e^y$ and thus $$ \begin{align*} \lim_{x \to \infty} \frac{\ln^r x}{x^p} &= \lim_{e^y \to \infty} \frac{y^r}{e^{py}} & e^{px} \geq e^x \text{ as } p> 1 \text{ and } x \geq 0\\ &\leq \lim_{y \to \infty} \frac{y^r}{e^y}\\ &= 0. \end{align*} $$

Step 2: Now I can prove that under the same hypothesis for $p$ and $r$, $$\lim_{x \to 0} x^p \ln^r x = 0.$$ Proof: Let $y = \frac{1}{x}$, thus $x = \frac{1}{y}$ and $$ \begin{align*} \lim_{x \to 0} x^p \ln^r x &= \lim_{\frac1y \to 0} \frac{1}{y^p} \cdot \ln \left( \frac{1}{y} \right)^r\\ &= \lim_{y \to \infty} \frac{- \ln^r y}{y^p}\\ &= 0, \end{align*} $$ which concludes.

I took $p > 1$ and $r \in \mathbb{N}$ because it is only what I needed for my exercise, but if you feel like editing and proving this for all $p. r > 0$, go ahead.

Answer 1

First let us examine the following. Given \begin{equation} l > 0 \end{equation} we prove that \begin{equation} \lim_{x \to0} ln(x)x^l = 0. \end{equation} Doing this comes straight from L'Hospitals' rule. We put the limit in the indeterminate form and get the following. \begin{equation} \lim_{x \to \infty}\frac{ln(1/x)}{x^l}= \lim_{x \to \infty}\frac{-1}{x} *\frac{1}{lx^{l-1}}=\lim_{x \to \infty} \frac{-1}{lx^{l}} = 0 \end{equation}

Since \begin{equation} l>0 \end{equation}

We now have a very general statement. From here you can see that if you substitute l for p/r (p and r >0), you can put our new found limit to the power of r and still get 0 as your answer. More importantly, by distributivity, we get the wanted limit like so.

\begin{equation} 0=\lim_{x \to 0}(x^{\frac{p}{r}}ln(x))^r =\lim_{x→0}x^p(ln(x))^r \end{equation}

Hope this helps.

Answer 2

I think you would need to prove that $$\lim_{y \to - \infty} e^{py} y^r = 0$$ using Lo'Hopital's Rule.

But you could just as easily start the problem by using Lo'Hopital's Rule, by first writing the expression as $$\lim_{x \to 0} \frac{ (\ln x)^r}{x^{-p}}.$$ You would have to use Lo'Hopital's Rule $r$ times, but at the end you should get $$\lim_{x \to 0} \frac{ c}{x^{-p}}$$ where $c$ is a constant. Clearly, this last limit evaluates to $0$.