# Show that $f(x) = \begin{cases}x^{\alpha}\sin(x^{-\beta}),& 0<x\leq 1 \\ 0,& x=0. \end{cases}$ is of bounded variation.

Define $$f(x) = \begin{cases}x^{\alpha}\sin(x^{-\beta}),& 0 Show that $$f$$ is of bounded variation on $$[0,1]$$ iff $$\alpha > \beta$$.

Here is a solution:

I didn't understand the following implication:

"... $$f$$ is of bounded variation on $$[0,x_{0}]$$, hence of bounded variation on $$[0,1]$$."

Could someone explain to me?