Let the function $f : \Bbb{R} \to \Bbb{R}$ be denoted by:
$$f(x) = \begin{cases} \dfrac13 \sin(3x) + x^4 \sin\dfrac{1}{x^3}, \quad & x \ne 0 \\ 0, & x = 0 \end{cases}$$
Prove that $f$ is uniformly continuous on $\Bbb{R}$
Should I prove this using the definition of uniform continuity? How can this be proved
Suppose there exists $M$ such that $|f'(t)| \leq M$ for all $t \in \mathbb{R}$. Then for all $x,y\in\mathbb{R}$, $$ f(x) - f(y) = f'(z)(x-y) $$ for some $z$ between $x$ and $y$. Therefore, $$ |f(x) - f(y)| \leq |f'(z)||x-y| \leq M |x-y| $$ (so $f$ is Lipschitz). Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{M}$. $\Box$
It remains to show that in this particular problem that $f'(x)$ is bounded, but that's a simple calculus problem.