f(x) = { x^2sin(1/x), if x =/= 0

0, if x=0}

a. find lim(x->0)f(x) and show that f(x) is continuous at x=0.

b, find f'(0) using the definition of the derivative at x=0:

f'(x)=lim(x->0) (f(x)-f(0)/x)

c. Show that lim(x->0)f'(x) does not exist. In particular, this means lim(x->0)f'(x) =/= f'(0), hence f'(x) is not continuous at x=0.

since |sinx| <= 1, x^2 sin(1/x) < x^2. So, since x^2 -> 0, so does x^2 sin(1/x)

x^2 sin(1/x) does not exist when x=0, but since f(0)=0, f is continuous.

[(0+h)^2 sin(1/(0+h)) - 0^2*sin(1/0)]/h
= [h^2 sin(1/h) - 0]/h
= h^2 ((1/h) - 1/3! (1/h^3) + ...)/h
= (h - h/3! + ...)/h
= 1 - 1/3! + ...
-> 5/6

using l'Hospital's rule

f'(x) = 2x sin(1/x) - cos(1/x)
2x sin(1/x) -> 0 but
cos(1/x) is undefined, somewhere between -1 and 1.

so, I guess 5/6 is as good a value as any.

a. To find the limit as x approaches 0 of f(x), we need to evaluate the function as x gets arbitrarily close to 0. Substituting x = 0 into the function gives f(0) = 0, so the value of the function at x = 0 is 0.

To check if f(x) is continuous at x = 0, we need to show that the limit of the function as x approaches 0 exists and is equal to f(0). In this case, we already found that f(0) = 0. Now we need to find the limit as x approaches 0 of f(x).

Let's consider the cases when x is not equal to 0. We have f(x) = x^2sin(1/x). As x approaches 0, sin(1/x) oscillates rapidly between -1 and 1, but the value of x^2 approaches 0. Therefore, the product x^2sin(1/x) approaches 0 as x approaches 0 from both sides.

So, the limit as x approaches 0 of f(x) is 0, which is equal to f(0). Therefore, f(x) is continuous at x = 0.

b. To find f'(0) using the definition of the derivative, we need to compute the limit as x approaches 0 of the expression (f(x) - f(0))/x.

We already know that f(0) = 0. Now, let's substitute these values into the expression to get:

f'(x) = limit as x approaches 0 of (f(x) - f(0))/x
= limit as x approaches 0 of (f(x))/x

For x not equal to 0, we have f(x)/x = x*sin(1/x). As x approaches 0, the factor x in the numerator approaches 0, while sin(1/x) oscillates rapidly between -1 and 1. Therefore, the limit of x*sin(1/x) as x approaches 0 does not exist.

So, f'(0) does not exist using the definition of the derivative at x = 0.

c. To show that the limit as x approaches 0 of f'(x) does not exist, we need to evaluate the expression f'(x) and find its limit as x approaches 0.

From part b, we found that for x not equal to 0, f'(x) = x*sin(1/x)/x = sin(1/x). As x approaches 0, sin(1/x) oscillates rapidly between -1 and 1, and the limit of sin(1/x) as x approaches 0 does not exist.

Therefore, the limit as x approaches 0 of f'(x) does not exist. In particular, this means that lim(x->0) f'(x) is not equal to f'(0), hence f'(x) is not continuous at x = 0.