Proof of limit of $x^3\ln x$ as $x$ goes to 0 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?
 A: 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 $\epsilon$.
A: 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.
A: $$\lim_{x \to 0^{+}} \left(x^3\ln (x)\right)=$$
$$\lim_{x \to 0^{+}} \left(\frac{\ln(x)}{\frac{1}{x^3}}\right)=$$
$$\lim_{x \to 0^{+}} \left(\frac{\frac{d}{dx}\ln(x)}{\frac{d}{dx}\left(\frac{1}{x^3}\right)}\right)=$$
$$\lim_{x \to 0^{+}} \left(\frac{\frac{1}{x}}{-\frac{3}{x^4}}\right)=$$
$$\lim_{x \to 0^{+}} \left(-\frac{x^3}{3}\right)=-\frac{0^3}{3}=0$$
