$f(0)=0,\;\;f(x)=e^{-2/x}\sin\left(e^{1/x}\right)$, is $f$ bounded variation on [0,1]?

Question

Let $$f(0)=0,\;\;f(x)=e^{-2/x}\sin\left(e^{1/x}\right),$$ is $f$ bounded variation on $[0,1]$?

Here is my thinking:

Since $f$ is differentiable on $(0,1]$ and continuous on $[0,1]$

If $f^\prime$ is bounded, we can use mean value theorem to prove it.

Answer 1

By taking the derivative we have $f'(x)=2y\sin(1/y)-\cos(1/y)$. Since we are over $[0,1]$, this is bounded by $2+1=3$. So we can assert bounded variation via mean value theorem.