Question

\[\lim \to infinity = \frac{ \sqrt{n^2+1+1} }{ 2n+3}\]

Answer 1

\[\lim_{ \rightarrow \infty}\frac{ \sqrt{n^2+1+1} }{ 2n+3} \]

Answer 2

1/3

Answer 3

\[\sqrt{n^2+1+1} = n+1+1 ??\]

Answer 4

Mmmmm I think you can do it this way.
Hopefully I'm not taking for granted the fact that I multiplied by 1/sqrt(n^2) which might cause an issue with the sign.. but I dunno.

No that little square root trick doesn't work pabek, I can explain it if you don't understand why c: