Let f (x) = x * sin (1 / x), X not equal to 0; when x equals 0, f (x) = 0, then why is the function continuous but not differentiable at x = 0 Or continuously differentiable, but how to prove it

Let f (x) = x * sin (1 / x), X not equal to 0; when x equals 0, f (x) = 0, then why is the function continuous but not differentiable at x = 0 Or continuously differentiable, but how to prove it

When x tends to 0, f (x) tends to 0, so it is continuous. The derivative is also defined. The derivative is 0, so it can be derived. But the derivative is not continuous

Let f (x) =, {(1 / x) * sin π x, X not equal to 0, a, x = 0, continuous at x = 0, find the value of A

lim(x-->0)f(x)
=lim(x-->0)π*(sinπx)/(πx)
=πlim(x-->0)(sinπx)/(πx)
=π
∵ f (x) is continuous at x = 0
∴lim(x-->0)f(x)=f(0)=π
∴a=π

Let f (x) = sin (1 / x) * (x ^ n) (when x is not equal to 0) f (x) = 0 (when x = 0) ask if n satisfies what condition, f (x) has a continuous derivative at x = 0

When n > 0, f (x) is continuous at x = 0, LIM (x - > 0) x ^ n * sin (1 / x) = 0, the product of bounded function and infinitesimal is infinitesimal. When n > 1, f (x) is differentiable at x = 0, and F '(0) = 0lim (x - > 0) [x ^ n * sin (1 / x) - 0] / x = 0. When x is not equal to 0, f' (x) = n x ^ (n