L'Hopital's rule questions, will fan/medal!

Something is seriously throwing my off. The first function is Lim as x goes to infinity ((5-x^2)(8+3x^2))/(5x^4-16)

I got my derivative as (-2x)(6x)/20x^3= infinity/infinity

then I got f'' as (-2)(6)/60x^2 = -12/infinity or 0. That isn't really checking out though.

Thank You!!

Question

L'Hopital's rule questions, will fan/medal!

Something is seriously throwing my off. The first function is Lim as x goes to infinity ((5-x^2)(8+3x^2))/(5x^4-16) 

I got my derivative as (-2x)(6x)/20x^3= infinity/infinity

then I got f'' as (-2)(6)/60x^2 = -12/infinity or 0. That isn't really checking out though. 

Thank You!!

anonymous · Answer

that numerator is a product of two binomials. Either multiply them, then take the derivative, or use the product rule.

tumblewolf · Answer

So Foil them out then try the process over again?

tumblewolf · Answer

Would the FOILED form be40+15x^2-8x^2+3x^4  ?

anonymous · Answer

I didn't check the whole thing, but something's off. The highest term should be -3x^4

tumblewolf · Answer

I get infinity/120 as  final

anonymous · Answer

ok. let's cut a lot of the extra stuff.

when you do l'hospital's rule, only the highest order terms matter. When you foil the numerator the highest term is $-3x^4$.  For the denominator the highest terms is $5x^4$. I'll leave it to you to include the lesser terms for your work.

Start by taking this limit
$$\lim_{x \rightarrow \infty}\frac{ -3x^4 }{ 5x^4 }$$

tumblewolf · Answer

infinity over infinity

tumblewolf · Answer

could you jsut pull the constants and get -3/5?

anonymous · Answer

yes you can

anonymous · Answer

that's pretty much l'hospital's rule for rational functions. If you keep taking the derivatives you end up with the constants.