I'm looking for help with a proof that I still cannot figure it out. Here is the statement:
"If $f:\mathbb{R} \rightarrow \mathbb{R}$ is $C^{1}$ and $\sup_{x \in \mathbb{R}}|f'(x)| = \infty$ , prove that $f$ cannot be uniformly continuous."
I know that if $f$ is $C^{1}$ and $\sup_{x \in \mathbb{R}}|f'(x)| = \infty$, $f$ cannot be Lipschitz. So I tried to prove that if $f$ is uniformly continuous $\Rightarrow$ $f$ is Lipschitz, because if it were true, if $f$ is NOT Lipschitz would mean $f$ is NOT uniformly continuous, and the proof would be resolved.
However, uniform continuity does not imply Lipschitz.
The statement is not true. Consider the function $$f(x)=\frac{\sin(x^5)}{x} \,.$$ with $f(0)=0$. Then, it is easy to see that $f \in C^1$ and since $f$ is vanishing at infinity it is uniformly continuous.
But $$f'(x)=\frac{5x^5\cos(x^5)-\sin(x^5)}{x^2}$$is clearly unbounded as $$f'(\sqrt[5]{2n\pi})=5\cdot (2n\pi)^\frac{3}{5}$$
Note If $\lim_{x \to \infty} |f'(x)|=\infty$ as Ted Shifrin suggested, then the claim follows immediately from the Mean Value Theorem. Indeed, if you assume that $f$ is uniformly continuous, then there exists some $\delta$ such that $|x-y|< \delta$ implies $|f(x)-f(y)|<1$. Apply the Mean value theorem on $(z,z+\frac{\delta}{2})$ and take the limit as $z \to \infty$ to get a contradiction.