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# Understand the problem

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 $\lim_{x_n \to \infty}\frac{f(x_n)}{|x_n|}\leq 10000$.
##### Source of the problem
TIFR 2019 GS Part A, Problem 14
Analysis
Moderate
##### Suggested Book
Real analysis Bartle and Sherbert

# Start with hints

Do you really need a hint? Try it first!

for sure option 1) is not true any $f(x)=|x|$ will give us the counter.

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 $x \not= 0$. What does it say about 3) and 4)?
This rules out (3) and (4) right away.
So (2) is the only option left.

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