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!!

Answer 1

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

Answer 2

So Foil them out then try the process over again?

Answer 3

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

Answer 4

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

Answer 5

I get infinity/120 as final

Answer 6

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
\[\lim_{x \rightarrow \infty}\frac{ -3x^4 }{ 5x^4 }\]

Answer 7

infinity over infinity

Answer 8

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

Answer 9

yes you can

Answer 10

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