$\lim_{n\to\infty}n^2(\int_{-1/n}^0u(x-s)ds -\int_0^{1/n}u(x-s)ds)$ where $u(x)$ an infinitely differentiable function on R

Question

$$\lim_{n\to\infty}n^2(\int_{-1/n}^0u(x-s)ds -\int_0^{1/n}u(x-s)ds)$$ where $u(x)$ an infinitely differentiable function on R.

Answer 1

$$ \int_{-1/n}^0 u(x-s)ds=\int_{0}^{1/n}u(x+s)ds. $$ Then $$ \lim_{n\to\infty}n^2(\int_{-1/n}^0u(x-s)ds -\int_0^{1/n}u(x-s)ds)\\ =\lim_{n\to\infty}\frac{\int_{0}^{1/n}[u(x+s)-u(x-s)]ds}{1/n^2}\\ \overset{t=1/n}{=} \lim_{t\to 0}\frac{\int_0^{t}[u(x+s)-u(x-s)]ds}{t^2}\\ \overset{L'Hospital}{=}\lim_{t\to 0}\frac{u(x+t)-u(x-t)}{2t}\\ =u'(x). $$