What's the limit as x approaches infinity, sqrt(x^4-2x^3-2)/(10x^2+9)

Question

jamesj · Answer

Divide top and bottom by x^2, then ...

anonymous · Answer

Since the top and bottom both go to infinity, you can use l'Hopital's rule!
$$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}$$

jamesj · Answer

In this case, it's liking shooting a fly with the gun of l'Hopital's rule.  Not necessary.

Follow my procedure and you see that the term becomes

sqrt(1-2/x-2/x^4)/(10+9/x^2)

Now what is the limit of that as x -> infinity?

anonymous · Answer

1/10?

anonymous · Answer

You can't divide the numerator by $x^2$ because it's inside a square root.

anonymous · Answer

oh ok...

jamesj · Answer

@Buck: yes, exactly

@Dalvoron: yes you can, notice that in doing so the term in the sqrt is divided by x^4.

anonymous · Answer

Touche.

jamesj · Answer

1/x^2 . sqrt(x^4-2x^3-2) = sqrt(1/x^4 . (x^4-2x^3-2) ) = ....

anonymous · Answer

so what if there wasn't a denominator? would it still apply?