Let f : [0, π] → R be defined byf (x) =(0 if x = 0,x sin 1x − 1x cos 1x if x̸ = 0.Is f continuous?

To determine if the function f is continuous, we need to check if the limit of the function as x approaches 0 is equal to the value of the function at x=0.

The function is defined differently for x=0 and x≠0, so we need to consider these cases separately.

For x=0, the function is defined as f(x)=0.

For x≠0, the function is defined as f(x)=x sin(1/x) - (1/x) cos(1/x).

We need to find the limit of this function as x approaches 0.

The limit of the first term, x sin(1/x), as x approaches 0 is 0. This is because the limit of x as x approaches 0 is 0, and the sin(1/x) term is bounded between -1 and 1.

The limit of the second term, -(1/x) cos(1/x), as x approaches 0 is more complicated. However, we can use the squeeze theorem to show that this limit is also 0. The cos(1/x) term is bounded between -1 and 1, so the entire term is bounded between -1/x and 1/x. As x approaches 0, both -1/x and 1/x approach infinity. Therefore, by the squeeze theorem, the limit of -(1/x) cos(1/x) as x approaches 0 is 0.

Therefore, the limit of the function as x approaches 0 is 0, which is equal to the value of the function at x=0.

So, the function f is continuous.