Determine whether f'(0) exists. f(x)=x^(2) sin 1/x if x is not equal to 0, 0 if x=0

1 Answers

Best Answer

Recall that:
f'(a)= lim_(x -> a) (f(x)-f(a))/(x-a)
f'(0)= lim_(x -> 0) (x^(2) sin ((1)/(x))-0)/(x-0)
f'(0)= lim_(x -> 0)x sin ((1)/(x))
Squeeze theorem
If
h(x) <= f(x) <= g(x)
And
lim_(x -> a)h(x)=L
And
lim_(x -> a)g(x)=L
Then
lim_(x -> a)f(x)=L
Since -1 <= sin ((1)/(x)) <= 1
Therefore -x <= x sin ((1)/(x)) <= x
Since
lim_(x -> 0)-x=0
And
lim_(x -> 0)x=0
By Squeeze Theorem, We have
lim_(x -> 0)x sin ((1)/(x))=0
Result:
f'(0)=0