# Limit of $\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}$

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.

• Do you realise the bracketed fraction tends to $\frac12$? Nov 2, 2016 at 12:55
• Hint for an alternative approach: First determine $\lim\limits_{n \rightarrow \infty} \ln \left( \frac{n^2 + 2}{2n^2 + 1} \right)^{n^2}$. Nov 2, 2016 at 12:57
• Notably, $(1/2)^\infty$ is not an indeterminate form Nov 2, 2016 at 12:57
• @MichaelJoyce: That is complicating the problem unnecessarily. Nov 2, 2016 at 12:58
• Right. To clarify I will just add that in this case you can use L'Hopital if you get an indeterminate like $0^\infty$. You don't have that here. So no L'Hopital. Nov 2, 2016 at 13:00

## 1 Answer

Hints:

Observe that

$$\frac{n^2+2}{2n^2+1}\xrightarrow[n\to\infty]{}\frac12\implies \exists\;N\in\Bbb N\;\;s.t.\;\;n>N\implies\frac12-0.01<\frac{n^2+2}{2n^2+1}<\frac12+0.01$$

Can you see now how to use the squeeze theorem?

• For a slightly easier lower bound, you can use $0$. This way you only need to worry about modifying the upper bound. Nov 2, 2016 at 12:59