How to prove that limit is equal to zero

Question

How to prove that:

$$\lim_{\epsilon\rightarrow 0}(-\log(-x) \phi(x)|_{-\infty}^{-\epsilon} -\log(x) \phi(x)|_{\epsilon}^{+\infty})$$ where $\phi(x) $ is any test function

is equal to $0$.

It seems pretty obvious, but I need formal proof. Thanks

Answer 1

Since $\phi$ is a test function, it has compact support and the limit becomes $\lim_{\epsilon \to 0} \left( \log \epsilon (\phi(\epsilon) - \phi(-\epsilon) ) \right) = \lim_{\epsilon \to 0} \left( (\epsilon \log \epsilon) { (\phi(\epsilon) - \phi(-\epsilon) ) \over \epsilon} \right)$.

Since $\phi$ is smooth, it is differentiable at $0$ hence ${ (\phi(\epsilon) - \phi(-\epsilon) ) \over \epsilon}$ is bounded for small $\epsilon$, and since $\lim_{\epsilon \to 0} ( \epsilon \log \epsilon) = 0$, we have the desired result.