(1) Show that f(x)=sin(1/x) is a continuous function on (0, 1].

(2) Prove that f(x)=sin(1/x) is not uniformly continuous on (0, 1].
Please give as much detail as possible.

(1) I'd use composition of functions for this.
g : (0, 1] → R via g(x) = 1/x and h : R → R via h(x) = sin x are continuous.
Hence, f(x) = (h o g)(x) = sin(1/x) is continuous on (0, 1].
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(2) This is best demonstrated with sequences.

Let {x(n)} = {1/(π/2 + 2πn)} and {y(n)} = {1/(2πn)} be sequence in (0, 1]..
So, |x(n) - y(n)| → 0, but |f(x(n)) - f(y(n))| = |1 - 0| = 1 does not converge to 0 as n→∞.

(So, for ε = 1, we have |f(x(n)) - f(y(n))| ≥ ε.)

Hence, f is not uniformly continuous on (0, 1].
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I hope this helps!

|f(x(n)) - f(y(n))| = |sin(π/2 + 2πn)) - sin(2πn)| = |1 - 0| = 1.

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