friends! I read on a book that, for $\alpha>1$, "being $g$ continuous in 0 [really $g$ is continuous in $[0,1]$, if it were useful to know] and approaching the extremes of the integral 0 for $n\to +\infty$ we have
$\int_{a/n^{\alpha}}^{n^{1-\alpha}} g(t) dt\sim g(0)(n^{1-\alpha}-a/n^{\alpha})$ for $n\to +\infty$."
Let $G$ be a primitive of $g$ and let's define $h(n)=\frac{n-a}{n^{\alpha}}$ and $f(n)=\frac{a}{n^{\alpha}}$; I would say that the formula above is equivalent to
$\lim_{n\to +\infty}\frac{G(n^{1-\alpha})-G(\frac{a}{n^{\alpha}}) }{\frac{n-a}{n^{\alpha}}} =\lim_{n\to +\infty}\frac{G(f(n)+h(n))-G(f(n))}{h(n)}=g(0) .$
In the case $f(n)$ would be a constant $c$ instead, I would immediately understand that
$\lim_{n\to +\infty}\frac{G(c+h(n))-G(c)}{h(n)}=\lim_{h\to 0^{+}}\frac{G(c+h)-G(c)}{h}=g(c)$
and, for $c\to 0^{+}$, $g(c)\to g(0)$. It's also clear to me that $\lim_{n\to +\infty}f(n)=0$. My problem is that both $f(n)$ and $h(n)$ "simultaneously" approach $0^{+}$, for $n\to +\infty$ and I am not sure whether it is possible to "decompose" the limit making $h(n)$ approach $0$ first and then doing the same with $f(n)$...
Would anybody help me to understand how this kind of limit is handled? Thank you so much!!!