This is a MCQ question, consider a function $f: \Bbb R \to \Bbb R$ s.t $|f(x)-f(y)|\leq 4321|x-y|$. Choose the correct option:
1) $f$ is always diffrentiable.
2) there exist atleast one such $f$ continuous but non-differentiable at exactly $2018$ points and $\lim_{x \to \pm \infty}\frac{f(x)}{|x|}=2018$
3) there exist atleast one such $f$ continuous s.t $\lim_{x \to \pm \infty}\frac{f(x)}{|x|}=\infty$
4) It is not possible to find a sequence of reals $\{x_n\}$ diverging to infinity s.t $\lim_{x_n \to \infty}\frac{f(x_n)}{|x_n|}\leq 10000$.
Now for sure option 1) is not true any $f(x)=1 $ forall $x>1$ and $f(x)=0$ forall $x\leq1$ will give us the counter but what are logics for the rest.
Best Answer
Observe that with $y = 0$ have $$|f(x)| \le |f(x) - f(0)| + |f(0)| \le 4321|x| + |f(0)|$$ so that $$ \frac{|f(x)|}{|x|} \le 4321 + \frac{|f(0)|}{|x|} $$ whenever $x \not= 0$.
This rules out (3) and (4) right away. The simple example $f(x) = |x|$ rules out (1).
So (2) is the only option left.