Show that f(x) defined on [0,infinity) R to is continuous at 0.

Question

$$f(x)=\begin{cases} 0 &:& x=0\\[1ex]x^{\frac{2}{3}} \log x&:&\text{otherwise}\end{cases}$$

My thoughts: $\log x$ behaves weird in $(0;1)$ and will not be bounded for any $ x \in V_{\delta} (0) , \delta >0 $

Now is $ x^{\frac{2}{3}}$ small enough to keep $\log x$ in control? How can I show this? Please help. I'm looking for a proof using the definition of continuity.

Answer 1

hint: $x^{\frac{2}{3}}\log x = \dfrac{\log x}{x^{-\frac{2}{3}}}$, and proceed to use L'hospitale rule to find the limit as $x \to 0^{+}$.

Answer 2

It is a basic fact of calculus that $$\lim_{x\to0+}x^\alpha\>\log x=0\qquad(\alpha>0)\ .\tag{1}$$ This immediately shows that your function is continuous at $0$.

There are various ways to prove $(1)$, depending on your toolbox. If you don't want to use Hôpital's rule you have to resort to $$\lim_{t\to\infty}{e^t\over t}=\infty$$ or similar. The most basic fact in this regard is $2^n>n$, which can be proven using Cantor's diagonal principle: The set $\{1,2,\ldots,2^n\}$ contains the $n+1$ numbers $1$, $2$, $\ldots$, $2^n$.