I want to evaluate the following integral using change of variables but I'm not sure if what I'm doing is correct. It 'feels' like I should get the answer $I = 2 u(0)$.
$$ I = \lim_{\varepsilon \to 0} \int_{-\varepsilon}^{\varepsilon} \frac{1}{\varepsilon} u(x) dx. $$
Ok so here is what I'm doing. Let $\hat x = x/\varepsilon$ and $\hat u(\hat x) = u(x)$. We have $dx = \varepsilon d\hat x$ and when $x=\pm \varepsilon$ we have $\hat x = \pm 1$. So this means that
$$ \begin{align} I & = \lim_{\varepsilon \to 0} \int_{-\varepsilon}^{\varepsilon} \frac{1}{\varepsilon} u(x) dx \\ & = \lim_{\varepsilon \to 0} \int_{-1}^{1} \frac{1}{\varepsilon} \hat u(\hat x) \varepsilon d\hat x \\ & = \lim_{\varepsilon \to 0} \int_{-1}^{1} \hat u(\hat x) d\hat x. \end{align} $$ The limits no longer depends on $\varepsilon$ so we can write \begin{align} I & = \int_{-1}^{1} \lim_{\varepsilon \to 0} \hat u(\hat x) d\hat x \\ & = \int_{-1}^{1} \lim_{\varepsilon \to 0} u(\varepsilon \hat x) d\hat x \\ & = \int_{-1}^{1}u(0) d\hat x \\ & = 2 u(0). \end{align}
Is this all correct and are all the steps fully rigorous?