Uniform continuity contradiction

Question

Question, I am learning uniform continuity and I saw a sentence which I am not sure about it ( since I can not find anything on it online ).
If the derivative is not bounded, and it limit on absolute value goes to infinity, so it is not uniform continuity.
Is the sentence good? or am I missing something?

Answer 1

This sentence is strange. It contains two assumptions: $f'$ is unbounded and $\lim_{x\to\infty}|f'(x)|=\infty$. But the second assumption implies the first one, and therefore the first one is not needed.

And the statement is true. Indeed, if $f$ was uniformly continuous, then, for every $\varepsilon>0$, there would exist a $\delta>0$ such that, for all $x,y$ such that $|x-y|<\delta$, $|f(x)-f(y)|<\varepsilon$. Since $\lim_{x\to\infty}|f'(x)|=\infty$, there exists $M>0$ such that for all $x>M$, $f'(x)>\frac{2\varepsilon}\delta$. Choose an arbitrary $x>M$ and apply the mean value theorem:there is some $c\in\left(x,x+\frac\delta2\right)$ such that$$\frac{f\left(x+\frac\delta2\right)-f(x)}{\left(x+\frac\delta2\right)-x}\left(=\frac{f\left(x+\frac\delta2\right)-f(x)}{\frac\delta2}\right)=f'(c)$$and then$$\left|f\left(x+\frac{\delta}{2} \right)-f(x)\right|=\bigl|f'(c)\bigr|\frac{\delta}{2}>\varepsilon.$$