Let $f(x) = \sqrt x \cos\frac{1}{x}$. Show $f$ is uniformly continuous on $(0, \infty)$.
Now as I recall, we've learned in class that if $f'(x)$ is bounded therefore $f$ is uniformly continuous.
Indeed,
$$ f'(x) = ... = \frac{1}{\sqrt x} \left( \frac{1}{2}\cos\frac{1}{x} + \frac{1}{x} + \sin\frac{1}{x} \right) $$
Boundesness:
$$\left|f'(x)\right| = \frac{1}{\sqrt x} \left( \frac{1}{2} \left|\cos\frac{1}{x}\right| + \frac{1}{x} \left|\frac{1}{x}\right| \right) \le \frac{1}{\sqrt x}(\frac{1}{2}+ \frac{1}{x}) \le \frac{3}{2}$$
My Question:
Is this sufficent? Because I've seen a solution for this excersise which using another approach:
By Cantor's theorem the funciton is uniformly continuous on $[0,1]$
And for $(1, \infty)$ we are showing $f$ is Lipschitz (and then proving uniform continuous directly by definition)
Isn't it useless if you can just claim $f$ is uniformly continuous by the derivative boundesness? Or am I wrong and my simple proof falls somewhere?
Not good. Why would $$\frac{1}{\sqrt x}(\frac{1}{2}+ \frac{1}{x}) \le \frac{3}{2}$$ be even remotely true? For small values of $x$, the statement is completely false...