Let $a_n$ be the $n$-th term of an arithmetic progression with the initial term $a_1=2$ and with the common difference $5$. That is, $a_n=2+5(n-1)$. Evaluate $\lim_{n\to\infty}n\left(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}\right)$.
This is what i asked solve, idk, does it because of my temperature,but i cant solve it more half a hour, what i actually did, just replace $a_n$ in $\lim$, that is
$(25n^2-30n+12)$ its all under root $-$ $(25n^2-30n+6)$ and this one under root
my apologizes if i cant see easy step, but could someone say me what i have do next?
HINT : $$\begin{align}n\left(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}\right)&=n\left(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}\right)\cdot\frac{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}\\&=\frac{n\left(a_n^2+3-(a_n^2-3)\right)}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}\\&=\frac{6n}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}\end{align}$$