Question

Which of these limits can L'Hopital's rule be applied to?
lim x->0 (4cos^2x)/x
lim x->0 e^(x^2)/x^2
lim x-> infinity sinx^2/x^2
lim x-> infinity (x^2-1)/(x+2)
lim x-> -2 (x^2-1)/(x+2)
Thank you so much for your time and help!

Answer 1

Basically there are five limits here, all produce the indeterminate form just by substituting, but four can be solved through other means. One requires L'Hopital, but I'm just not sure which. There are so many that it's daunting and I'm unsure how to solve most of them. But I'll try to get started, and I would love some help!!

Answer 2

All 5 can be applied with L'Hopital's Rule... Since we got a infinity/infinity situation for all 5.

Answer 3

They are all indeterminite :).

Answer 4

Just left alone, that is.

Answer 5

Right, I know that the rule will work for all of them... But the question is multiple choice, and there is only one answer... so they are asking for which limit is the only solution L Hopital, which one uses it as a last resort. :)

Answer 6

lim x-> -2 (x^2-1)/(x+2)


doesnt work here right ?

Answer 7

I believe it would be the second one.

Answer 8

I dont see quite how it works here too :

lim x->0 (4cos^2x)/x
lim x->0 e^(x^2)/x^2

Answer 9

Since the the numerator and denominator both got to 0/0 when you substitute.

Answer 10

oh thank you for the observation ganeshie, let me try solving them with the rule.

Answer 11

cos(0) = 1
e^0 = 1

Answer 12

Ahhh yea, I just realized that xD thanks Ganeshie8.

Answer 13

:)

Answer 14

Its definitely the 4th then... infinity - constant/ infinity minus constant

Answer 15

yep only 3rd and 4th i think

Answer 16

Noooo 4th one is okay.

Answer 17

Oh wait a second, 3rd wont work

Answer 18

Only 4th works

Answer 19

Limit for the 5th one is 3/4 i believe.

Answer 20

you are saying that only the fourth one will use L'Hopital?

Answer 21

Yes. That is exactly what we are saying.

Answer 22

I'm still a little confused, can you explain why the fourth one is the answer and why the others won't "work?"

Answer 23

\( \lim \limits_{x \to 0} \dfrac{4\cos^2x}{x} = \dfrac{4\cos^2 (0)}{0} = \dfrac{4*1}{0} = \dfrac{4}{0} \)

Answer 24

its not of form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), so it doesnt satisfy the hypothesis requirement fo rapplying L'hospital's rule.

Answer 25

oh, I had forgotten about the required indeterminate forms! Thanks so much for clearing this up!

Answer 26

np :) review this if u have time :
http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

Answer 27

Always remember, if something is divided by zero in the calculus world, it doesn't always mean it is 0 ;P

Just to be friendly! WELCOME TO THE CALCULUS UNIVERSEEEEE WOOOOOOH! lol

Answer 28

Haha, thanks for your help, you guys! :D I'll check that out, ganeshie, sounds helpful. Question closing!

Answer 29

Anytime :)