Question

lim h->infinity, (3x^2 +4x)^3/(4x^3-3)^2,,,can somebody help me figure this out? im staring at it for half an hour now

Answer 1

Is this the right question though?

Answer 2

yea, it is

Answer 3

no

Answer 4

it is x tends to infinity

Answer 5

u r photo is nice anyss

Answer 6

i became ur first fan

Answer 7

..............

Answer 8

actually ur right x tends to infinity,, can u help me?

Answer 9

OMH aravindG you are so creepy

Answer 10

lols

Answer 11

see ya

Answer 12

i think ktklown got the answer else i will answer later

Answer 13

anyss if i remember correctly you have to use l'hopital's rule which means differentiate the numerator and denominator and re-evaluate

if the num and denom are still both infinity, differentiate again. keep doing this until you get a determinate answer.

Answer 14

owh, ktklown, im trying now

Answer 15

why don't you take the limit directly

Answer 16

L'hopital rule is long because your denominator is not = 0

Answer 17

i think the limit directly ends up just being infinity over infinity

Answer 18

sorry i thout your lim as x goes to 0

Answer 19

i was just dumb

Answer 20

should i take out the highest power of x then?

Answer 21

i'm working this through with mathematica and it's a real pain, you have to take the derivative 6 times of both the numerator and denominator before you finally get the answer. that's a really obnoxious question.

Answer 22

the final answer is 27/16 but that would take hours to compute by hand

Answer 23

its the past year exam question actually,,im just trying it out,,and ur rite,,it is a real pain

Answer 24

i mean the number of times you'd have to use the chain rule, etc.... it is obnoxious. i hate math teachers, it's exactly this kind of question that made me hate math when i was your age

Answer 25

anyway in theory you keep just differentiating the numerator and denominator separately as long as they're both infinity or both 0

Answer 26

once you differentiate them for a 6th time they both turn into constants, and they divide out to 27/16

Answer 27

ohhhh,,,tanx btw

Answer 28

Multiply out the numerator and the denominator:
\[\lim_{x\rightarrow \infty} \frac{(3x^2+4x)^3}{(4x^3-3)^2}=\frac{(3x^2+4x)(3x^2+4x)(3x^2+4x)}{(4x^3-3)(4x^3-3)}\]
\[=\frac{27x^6+108x^5+144x^4+64x^3}{16x^6-24x^3+9}\]
Divide everything by the highest power of x (i.e. x^6):
\[\lim_{x\rightarrow \infty} \frac{27x^6+108x^5+144x^4+64x^3}{16x^6-24x^3+9} =\frac{27 + \frac{108}{x} +\frac{144}{x^2}+\frac{64}{x^3}}{16-\frac{24}{x^3}+\frac{9}{x^6}}\]
Since x is going to infinity, and we're looking for the limit, everything divided by some power of x becomes very very close to zero (so close that we may assume it is equal to zero):
\[\lim_{x\rightarrow \infty} \frac{27 + \frac{108}{x} +\frac{144}{x^2}+\frac{64}{x^3}}{16-\frac{24}{x^3}+\frac{9}{x^6}}=\frac{27+0+0+0}{16+0+0}=\frac{27}{16}\]