# 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.

# First look at the knowledge graph.

# 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,

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

**sandwich**)

**theorem**

##### Difficulty Level

7/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

# Start with hints

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 havef'(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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