Let \(f:[0,\infty)\to \mathbb{R}\) be defined by

\( f(x)=x^{2/3}logx\) for \(x>0 \) \(f(x)=0\) if \(x=0\)


A. f is discontinuous at x=0

B. f is continuous on \([0,\infty)\) but not uniformly continuous on \([0,\infty)\)

C. f is uniformly continuous on \([0,\infty)\)

D. f is not uniformly continuous of \([0,\infty\) but uniformly continuous of \((0,\infty)\)


First, we find whether \(f\) is continuous or not. \(f\) is “clearly” continuous on \((0,\infty)\).

Question is, what happens near \(x=0\)?

Take limit.

\(lim_{x\to 0}\frac{logx}{x^{-2/3}}\) is in \(\frac{-\infty}{\infty}\) form.

We use the L’Hospital Rule.

\(lim_{x\to 0}\frac{logx}{x^{-2/3}}\)=\(lim_{x\to 0}\frac{1/x}{-\frac{2}{3}x^{-2/3-1}}\)

=\(lim_{x\to 0}-3/2{x^{2/3}}\)


Okay! So we now know that \(f\) is at-least continuous.

Now we present the HINT s:

  • COntinuous function on a compact interval is uniformly continuous.
  • If derivative is bounded the by using mean-value theorem we can prove that f is Lipschitz, and therefore uniformly continuous.

For \(x>0\), \(f'(x)=\frac{2}{3}x^{-1/3}logx + x^{2/3}x^{-1}\) (By the formula for calculating derivatives of product).

After simplification,


We want the boundedness of \(f’\) and we don’t care about what happens close to zero as much. (Because we are going to use the first hint).

\(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}\) is in \(\frac{\infty}{\infty}\) form.

Once again, we use the L’Hospital rule.

\(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}=lim_{x\to \infty}\frac{2/x}{3/3 x^{-2/3}}\)

\(= lim_{x\to \infty} 2x^{-1/3}=0\)

What does this tell us? This tells us that \(f’\) is bounded on (say) \([1,\infty)\)

So \(f\) is uniformly continuous on \([1,\infty)\). Also, \(f\) is uniformly continuous on \([0,1]\).

So given \(\epsilon>0\) we have two “delta’s” for these two intervals. (which satisfies the condition for uniform continuous in these intervals respectively). Take the delta which is less. And this gives the uniform continuity for \(f\).