I trying to calculate  $$ \lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}.\tag{1}$$  Of course I cannot use theorem that if $a_n\to\infty$ then $\lim_{n\to\infty}(1+\frac{1}{a_n})^{a_n}=e,$ because in this particular situation we have $$\lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1-n^2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1}{\frac{(2n^2+1)}{(1-n^2)}}\big)^{n^2},$$ and $\lim_{n\to\infty}\frac{(2n^2+1)}{(1-n^2)}=-2.$  I would be grateful of any hints and comments.
The limit to solve is $$\lim_{n\to\infty}\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}$$ I tried to use L'Hôpital's rule but the derivatives are quite complex.
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