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# What are we learning ?

Competency in Focus: Limit and Differentiability of a function

This problem from I.S.I. B.Stat. Entrance 2017 , it is based on simple manipulations,  differentiation from first principles and squeeze theorem.

# Next understand the problem

Let f : R → R be a continuous function such that for any two real
numbers x and y, $|f(x)-f(y)| \le {|x-y|}^{201}$.
Then,
##### Source of the problem
I.S.I. B.Stat. Entrance 2017, UGA Problem 5
##### Key Competency
Differentiation from first principles and squeeze (or sandwichtheorem
7/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
Given that $|f(x)-f(y)| \le {|x-y|}^{201}$ for any two real number x and y . So we can replace x by y+h for any real number y and h .Then we have , $|f(y+h)-f(y)| \le {|(y+h)-y|}^{201} \implies |f(y+h)-f(y)| \le {|h|}^{201}$ . Now,  what can we use ?

See that we can write  $\frac{ |f(y+h)-f(y)|}{h} \le {|h|}^{200} \implies -{h}^{200} \le \frac {|f(y+h)-f(y)|}{h} \le h^{200}$ .  Now taking limit h tends to 0 on both sides gives what ?

 See that $\lim_{h \to 0} -{h}^{200}=\lim_{h \to 0} {h}^{200}=0$ and differentiation from first principles say’s $\lim_{h \to 0} \frac{f(y+h)-f(y)}{h} =f'(y)$ .
Now we know the the squeeze theorem, stated as follows   Let I be an interval having the point a as a limit point. Let g, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have
and also suppose that
Then So, what we have ?
f'(y)=0 for all y belongs to real . Thus integrating both sides we have f(y)=c ( constant ) for all y belongs to real . Therefore , f(101)=f(202)=f(200).

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