Let $f$ be odd real locally integrable function such that $$\lim_{x\rightarrow\infty}xf(x)=1$$
I need to show that $g_{\varepsilon}(x)=\varepsilon^{2}f'(\varepsilon x)$ has limit in sense of distributions and find the limit. First we get this:$$\left<g_{\varepsilon},\varphi\right>=\int_{-\infty}^{+\infty}\varepsilon^{2}f'(\varepsilon x)\varphi(x) \, dx=\varepsilon\int_{-\infty}^{+\infty}\varepsilon f'(\varepsilon x) \, \varphi(x) \, dx=-\varepsilon\int_{-\infty}^{+\infty}f(\varepsilon x) \, \varphi'(x) \, dx$$ with last equality provided by partial integration. I pretty much don't know how to proceed from here (because another p.i. brings me back to the beggining) so I'd be glad if someone can offer any hint.
Edit: if $f$ is odd, does it mean that $f'$ is even?