Problem 1. Suppose that f : R → R is a function that satisfies f (0) = 0 and f ′(0) = 0.Define the function g : R → R byg(x) ={f (x) · sin 1x if x 6 = 0,0 if x = 0.Prove that g is differentiable at 0 and g′(0) = 0

To prove that g is differentiable at 0 and g′(0) = 0, we need to show that the limit as x approaches 0 of [g(x) - g(0)] / x exists and equals 0.

Given that g(x) = f(x) * sin(1/x) for x ≠ 0 and g(0) = 0, we can substitute these into the limit:

lim (x->0) [g(x) - g(0)] / x = lim (x->0) [f(x) * sin(1/x) - 0] / x = lim (x->0) f(x) * sin(1/x) / x.

We know that f(0) = 0 and f′(0) = 0. This means that f(x) is small when x is close to 0. Therefore, we can say that f(x) is bounded by some M|x| for x close to 0, where M is a constant.

So, |f(x) * sin(1/x) / x| ≤ |M * sin(1/x)|.

As x approaches 0, sin(1/x) oscillates between -1 and 1, so |sin(1/x)| ≤ 1. Therefore, |M * sin(1/x)| ≤ |M|.

Now, we can apply the Squeeze theorem, which states that if a function h(x) is always less than or equal to g(x) and always greater than or equal to f(x), and the limits as x approaches a certain value c of f(x) and g(x) are equal, then the limit of h(x) as x approaches c is also equal to this common limit.

Applying the Squeeze theorem to our limit, we get:

0 ≤ |f(x) * sin(1/x) / x| ≤ |M|.

Taking the limit as x approaches 0, we get 0 ≤ lim (x->0) |f(x) * sin(1/x) / x| ≤ 0.

Therefore, by the Squeeze theorem, lim (x->0) f(x) * sin(1/x) / x = 0.

This shows that g is differentiable at 0 and g′(0) = 0.