Prove that $x^\frac{3}{2}\sin (\frac{1}{x})$ is a function of bounded variation for $x\in(0,1]$.

Question

I am studying for a test in measure theory. Please help with the following question:

Prove that $x^\frac{3}{2}\sin (\frac{1}{x})$ is a function of bounded variation for $x\in(0,1]$.

Answer 1

Hint: if $f$ is differentiable on $(0,1)$ then $\operatorname{Var}(f)=\int_0^1 |f'(x)|\mathrm{d}x$.

Answer 2

$$\int_{0}^{1}\left|\left(x^{3/2}\sin\left(1/x\right)\right)'\right|dx\leq\frac{3}{2}\int_{0}^{1}\left|\sqrt{x}\sin\left(\frac{1}{x}\right)\right|dx+\int_{0}^{1}\frac{\left|\cos\left(\frac{1}{x}\right)\right|}{\sqrt{x}}dx\leq $$ $$\leq\frac{3}{2}\int_{0}^{1}\sqrt{x}dx+\int_{0}^{1}\frac{1}{\sqrt{x}}dx=3. $$