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I am trying to find an $\epsilon$-$\delta$ proof of \begin{equation*} \lim_{x \to 0^{+}} x^3\ln x=0 \end{equation*} Is there a way to construct such a $\delta$ and not find it by educated guessing?

Emmet
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    You cannot get rid of that $\ln$ so easily, as $|\ln x|$ is not smaller than $|\ln (e^{-\epsilon})|$ when $|x| < e^{-\epsilon}$. –  May 23 '15 at 09:06
  • In these proofs I usually feel like overkill is fine. Choose something which works out simply. – Stanley May 23 '15 at 09:08
  • Using the $\epsilon - \delta$ argument, suppose $\epsilon > 0$, you have $|x|^3|\ln(x)| < \epsilon$. Though I thought about it for a bit, but could you do the following. We need to get some $\delta > 0$ so that $x \in (0,\delta)$ implies the above inequality. But if we can represent $x_0 = e^{-x_1}$ for some $x_1 > 0$, then we have $|e^{-x_1}|^3|ln(e^{-x_1})| = \frac{|-x_1|}{e^{3x_1}}$. If we pick $x_1 > 0$ sufficiently large, then we can get $\frac{|-x_1|}{e^{3x_1}} < \epsilon$. Just a thought, though my argument isn't rigorous, rather a sketch. – MathNewbie May 23 '15 at 09:20
  • @John You are right, I did a mistake. Let me modify my question. – Emmet May 23 '15 at 09:38
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    $\lim\limits_{x\to0^+}(x^3\ln{x}) = \lim\limits_{x\to0^+}(\ln{x^{x^3}}) = \ln(\lim\limits_{x\to0^+}{x^{x^3}}) = \ln1 = 0$ – barak manos May 23 '15 at 11:57
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    @barakmanos How do you know $x^{x^3} \to 1$ without first showing $x^3 \ln x \to 0$? – Erick Wong May 23 '15 at 16:32
  • @ErickWong: $x^3$ converges to $0$ "much faster" than $x$, hence $x^{x^3}$ converges to $1$. But you're right in the sense that this is not mathematical reasoning. – barak manos May 23 '15 at 19:23
  • @barakmanos In that case why not just say $x^3$ converges to $0$ "much faster" than $1/\ln x$ and deduce the original limit directly? – Erick Wong May 23 '15 at 23:25
  • @ErickWong: Because this one is not so obvious. – barak manos May 24 '15 at 07:48
  • @barakmanos 1) I disagree with the obviousness, on the grounds that most people have poor intuition about limits of type $0^0$ compared to $0/0$. 2) Is there some way that converting to $x^{x^3}$ helps towards an $\epsilon-\delta$ argument? I don't see it. – Erick Wong May 24 '15 at 08:32

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Setting $\ln(x) = -t$, we have $$\left \vert x^3 \ln(x) \right \vert = \left \vert -te^{-3t} \right \vert = \left \vert -\dfrac{t}{e^{3t}} \right \vert = \left \vert \dfrac{t}{e^{3t}}\right \vert < \left \vert \dfrac{t}{(3t)^2/2}\right \vert = \dfrac2{9t}$$ This will let you construct $\delta$ based on your choice of $\epsilon$.

Adhvaitha
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You can fall back on the definition of $\ln x$ to simplify things. For $0 < x < 1$ we have

$$ | \ln x | = \int_x^1 \frac{dt}{t} < \int_x^1 \frac{dt}{t^2} = \frac{1}{x} - 1, $$

so

$$ \left|x^3\ln x\right| < x^3 \left(\frac{1}{x} - 1\right) = x^2 - x^3 < x^2. $$

Now you just need to find a $\delta > 0$ such that $x^2 < \epsilon$ for $0 < x < \delta$, which doesn't require any guesswork.

Antonio Vargas
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$$\lim_{x \to 0^{+}} \left(x^3\ln (x)\right)=$$ $$\lim_{x \to 0^{+}} \left(\frac{\ln(x)}{\frac{1}{x^3}}\right)=$$ $$\lim_{x \to 0^{+}} \left(\frac{\frac{d}{dx}\ln(x)}{\frac{d}{dx}\left(\frac{1}{x^3}\right)}\right)=$$ $$\lim_{x \to 0^{+}} \left(\frac{\frac{1}{x}}{-\frac{3}{x^4}}\right)=$$ $$\lim_{x \to 0^{+}} \left(-\frac{x^3}{3}\right)=-\frac{0^3}{3}=0$$

Jan Eerland
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