Question

test the differentiability of the function at x=0 F(x)= {x*sin(1/x)if x≠0 and 0 if x=0

Answer 1

It is not differentiable. you can check that the function is continous.
Applyinf frist principle \[f \prime(0)=\lim_{h \rightarrow 0}(f(h)-f(0))/h=hsin(1/h)/h=\sin1/h\] sin1/h is undefined as h tends to 0 hence it is not differentiable.

Answer 2

but can't we say 1/h->infinity

Answer 3

Yeah since 1/h->infinty the limit sin(1/h) and subsequently f'(0) is undefined.

Answer 4

If you have x^a sin(1/x) it is differentiable at x=0 only for a>1 for an integer a am not sure for fractional a value.

Answer 5

The derivative exists at x=0 if the limit of the difference quotient
\[ \frac{f(h) - f(0)}{h} = \sin(1/h) \]
as h -> 0 exists.

But it doesn't exist because
\[ \lim_{h \rightarrow 0} \sin(1/h) \]
doesn't exist.

Instead, as h approaches 0, from either above or below, sin(1/h) oscillates faster and faster between -1 and 1.

If it had a limit, L say, then we could make sin(1/h) as close as we liked to L by making h sufficiently close to zero. But on the contrary, for any restriction of h close to zero, sin(1/h) assumes infinitely many values between -1 and 1.

Answer 6

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Answer 7

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Answer 8

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Answer 9

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Answer 10

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Answer 11

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Answer 12

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Answer 13

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Answer 14

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f(x) = x[x] ; \(0\le x < 2\)
= (x-1)x ; \(2 \le x \le

Answer 15

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