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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    $\begingroup$ Do you realise the bracketed fraction tends to $\frac12$? $\endgroup$ Nov 2, 2016 at 12:55
  • $\begingroup$ Hint for an alternative approach: First determine $\lim\limits_{n \rightarrow \infty} \ln \left( \frac{n^2 + 2}{2n^2 + 1} \right)^{n^2}$. $\endgroup$ Nov 2, 2016 at 12:57
  • $\begingroup$ Notably, $(1/2)^\infty$ is not an indeterminate form $\endgroup$ Nov 2, 2016 at 12:57
  • $\begingroup$ @MichaelJoyce: That is complicating the problem unnecessarily. $\endgroup$
    – Clayton
    Nov 2, 2016 at 12:58
  • $\begingroup$ 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. $\endgroup$ Nov 2, 2016 at 13:00

1 Answer 1



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?

  • 1
    $\begingroup$ For a slightly easier lower bound, you can use $0$. This way you only need to worry about modifying the upper bound. $\endgroup$
    – Clayton
    Nov 2, 2016 at 12:59

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