Question

Lim(x->inf) x^2/sqrt(x^4+1)

Answer 1

do you know l'hospital's rule?

Answer 2

Yes, but I'm not very good at applying it. :P

Answer 3

you have an indeterminate form...\[\frac{ \infty }{ \infty }\]
So you can take the derivative of the top and the bottom to see if it resolves...\[\lim_{x \rightarrow \infty}\frac{ x ^{2} }{ \sqrt{x ^{4}+1} }=\lim_{x \rightarrow \infty}\frac{ \frac{ d }{dx } \left( x ^{2} \right) }{ \frac{ d }{ dx } \left( \sqrt{x ^{4}+1} \right) }\]

Answer 4

if you get an indeterminate form from that, you can apply l'hospital's rule again, and again (if needed, as long as you want)

Answer 5

actually, L'Hopital's rule doesn't quite work directly in this case...

you could perhaps use the squeeze theorem...

\[\frac{ x^{2} }{ x^{2}+1 }<\frac{ x^{2} }{ \sqrt{x^{2}+1} }<\frac{ x^{2} }{ x^{2}-1 }\text{, }\forall x>1\]

you would use L'Hospital's rule twice on both ends (not the middle) and show that they each go to 1 and by the squeeze theorem, the middle must also go to 1 (as x goes to infinity)

Answer 6

oops, that should be x^4 inside the square root

Answer 7

you could also square it...

\[\left[ \lim_{x \rightarrow \infty}\frac{ x^{2} }{ \sqrt{x^{4}+1} } \right]^{2} = \lim_{x \rightarrow \infty}\left( \frac{ x^{2} }{ \sqrt{x^{4}+1} } \right)^{2}=\lim_{x \rightarrow \infty} \frac{ x^{4} }{ x^{4}+1 } =1\text{ using L'Hospital's rule 4 times}\]
then take the square root to get 1 (since both top and bottom of the original are positive)