Okay, so I'm trying to use a theorem that says if the derivative to the function has a max at the given interval then the function is uniformly continuous there.
I don't know how to continue my calculations to find the max value to the derivative.Here is the task and what I have done:
Find out if $$f(x)=x^2 \cos\left(\frac{1}{x}\right)$$ is uniformly continuous at $$I=(0,1)$$
My calculations:
$$f'(x)=2x \cos\left(\frac{1}{x}\right)+ \sin\left(\frac{1}{x}\right) \leq 2x \cos\left(\frac{1}{x}\right)+ \sin\left(\frac{1}{x}\right)$$
But here it stops for me, because the $0$ in the interval makes it hard for me to understand what to do with the expression. Does anyone have some tips? How do I find the max-value to the derivative?
Fixing the proof of the OP:
We have $$ f'(x)=2x \cos\left(\frac{1}{x}\right)+ \sin\left(\frac{1}{x}\right) $$ and hence $$ |f'(x)|\leq 2x \Big|\cos\left(\frac{1}{x}\right)\Big| + \Big|\sin\left(\frac{1}{x}\right)\Big|\le 3. $$ MVP provides that for every $x,y\in(0,1)$, there is a $\xi \in(x,y)\subset(0,1)$, such that $$ |f(x)-f(y)|=|f'(\xi)||x-y|\le 3|x-y| $$ and hence $f$ is uniformly continuous ($\delta=\varepsilon/3$).