I am aware that proving uniform contnuity can sometimes be easier using the Mean Value Theorem. Eventually it boils down to proving that the derivative is bounded. For illustration purposes:
Let $f: \mathbb R \to \mathbb R$ and ${\lim_{x \rightarrow \infty}f^\prime(x)}=0$
By definition that means $\exists c \in \mathbb R$ such that $|f^\prime(x)|<c$ and my question is then for which $x \in \mathbb R$ is the above valid.
I'm playing with the idea that it suffices if ${\lim_{x \rightarrow \infty}f^\prime(x)}=a$, where $a \in \mathbb R$ which would then mean our sequence $(f^\prime(x))_{x}$ is bounded. The thought that, however, seems to nullify that thought process is that we're in $\mathbb R$ and this may not hold for ${\lim_{x \rightarrow -\infty}f^\prime(x)}$.
Am I missing something? Any help would be greatly appreciated.
But there exist functions $f$ which are uniformly continuous but not having bounded derivative, so the strategy regarding bounded derivative technique sometimes does not work:
Find a $\phi\in C^{\infty}({\bf{R}})$ such that $\phi=1$ on $|x|\geq 1$ and $\phi=0$ on $|x|<\delta$ for small $\delta\in(0,1)$, then let $f(x)=\dfrac{\sin(x^{3})}{x}$ on $|x|>\delta/2$, $f(x)=0$ for $|x|\leq\delta/2$, and consider $\phi\cdot f$, then $\phi\cdot f:{\bf{R}}\rightarrow{\bf{R}}$ is uniformly continuous but for all $x>1$, \begin{align*} (\phi\cdot f)'(x)=3x\cos(x^{3})-\dfrac{\sin(x^{3})}{x^{2}}, \end{align*} so $\phi\cdot f$ has unbounded derivative.