[Math] Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ of bounded variation on $[-1,1]$

Question

Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ with $f(0)=0$ of bounded variation on $[-1,1]$?

I know that $g(x)=x\cos\left(\frac{1}{x}\right)$ and $h(x)=x^2\cos\left(\frac{1}{x^2}\right)$ are both NOT of bounded variation, so I'm guessing that $f$ isn't either…? I just don't know which partition to take, since the proofs of the ones for $g$ and $h$ involved expressing the Variation into a harmonic series, which diverges when the number of points in the partition increases, but I don't know how to do the same for $f$. Thanks!

Answer 1

Hint: A function with bounded derivative is Lipschitz.