# Whether f ′ ( 0 ) exits or not.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2.6, Problem 54E
To determine

Expert Solution

## Answer to Problem 54E

The function f(0) exists.

### Explanation of Solution

Squeeze theorem:

Suppose an inequality of functions g(x)f(x)h(x) in a interval around c and that limxcg(x)=L=limxch(x). Then limxcf(x)=L.

Definition:

A function f(x) is differentiable at point x, then f'(x)=limh0f(x+h)f(x)h.

Calculation:

Consider the function f(x)=x2sin(1x) when x0 and f(0)=0,

By definition, f'(x)=limh0f(x+h)f(x)h.

Substitute x=0 in the above equation,

f'(0)=limh0f(0+h)f(0)h=limh0f(h)f(0)h=limh0h2sin(1h)0h=limh0hsin(1h)

Since the value of sin(1h) lies between 1 to 1.

That is 1sin1h1.

Multiply throughout by h,

hhsin1hh

Thus limx0(h)=0 and limx0h=0.

By squeeze theorem,

limx0hsin(1h)=0

Therefore, the limit of f'(0) exist.

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