Question

L'Hopital's rule help

my functions are

lim as x apporaches 0 (4-x^2)(7+3x^2)/10x^4+28 and I got -3/10 and my lim

lim as x apporaches 0 e^2x-cos2x/x^2+9x my answer was 1/9

lim as x apporaches infinity 5x^3-4x-8/ 5x^4+8x+3 and I got 0

lim as x apporaches infinity 6x/lnx and I got 0

@agent0smith can you help?

Answer 1

HI!!

Answer 2

if you get \(\frac{0}{0}\) with polynomials, that means you can factor and cancel

Answer 3

I usually just keep taking derivatives until something works

Answer 4

did you try plugging in 0 first?

Answer 5

yes

Answer 6

what did you get? erase all the x's
it is not what you wrote

Answer 7

\[\frac{4\times 7}{28}=?\]

Answer 8

=1?

Answer 9

yes

Answer 10

you can't use l'hopital here, since it is not "indeterminate" it is just 1

Answer 11

how do you get to the just (4) (7) /28?

Answer 12

oh never mind on that part

Answer 13

The last three are the ones I really need help with

Answer 14

L'hopital's rule is easy. Take the derivative of the numerator, take the derivative of the denominator, then see if you can evaluate it.

Answer 15

My numbers just seem really weird

Answer 16

the second problem is indeterminate 0/0 so you can use l'hopital
2e^2x + 2sin2x/ 2x+9

Answer 17

now just put x=0

Answer 18

2+2/9?

Answer 19

sin0 is not 1

Answer 20

It would be cos(o) right?

Answer 21

for the third one you just use L'hopital twice and you'll end up in the form of
Lim x-> inf A/x which is 0

Answer 22

for the 4th one using L'hopital once we get

Lim x->inf 6/(1/x) = Lim x->inf 6x

Answer 23

This is not zero

Answer 24

\[\frac{ 6 }{ \frac{ 1 }{ x } }\]

Answer 25

is this not right?

Answer 26

yes and that becomes 6x

Answer 27

for the 2nd one the answer is 2/9

Answer 28

How do you determine the final answer for the 6x

Answer 29

What does 6x approach, when x approaches infinity?

Answer 30

infinity?

Answer 31

well done

Answer 32

Yes

Answer 33

Thank you both!