## suju101 3 years ago test the differentiability of the function at x=0 F(x)= {x*sin(1/x)if x≠0 and 0 if x=0

1. shankvee\

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.

2. suju101

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

3. shankvee\

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

4. shankvee\

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.

5. JamesJ

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.