Let f (x) = {(x ^ a) sin [1 / (x ^ b)], X not equal to 0.0, x = 0.} when a, B take what value, f (x) is continuous at x = 0?

Let f (x) = {(x ^ a) sin [1 / (x ^ b)], X not equal to 0.0, x = 0.} when a, B take what value, f (x) is continuous at x = 0?

If b > = 0, only a > 0 is needed, sin (1 / x ^ b) is a bounded function
If B0 is enough

When a is a value, f (x) = piecewise function -- X ^ 2-4 / X-2, X is not equal to 2; a, x = 2 -- continuous at x = 2

f（x）=(x^2-4)/(x-2)=(x+2)(x-2)/(x-2)=x+2 ,x≠2
=a ,x=2
Continuous at x = 2
When ∵ x = 2,
x+2=4
But the actual x = 2 cannot be replaced by X + 2,
If we want to be continuous, we can only supplement the position of function x + 2 with F (x) = A and X ≠ 2
That is, a = 4

When x is not equal to 0, f (x) = SiNx / X; when x = 0, f (x) = 1:1
Why is the n-th derivative of F (0) substituted by 0 after the derivation of F (x) = SiNx / X instead of the n-th derivative of F (0) = 1?

It's like finding the derivative of y = x at x = 1
If the number is substituted, it will become a constant, and then the derivative will be 0