Question

Limit help

Answer 1

\[\lim_{x \rightarrow 0} (4-2x^3+x)\div(8x+6x^3)\]

Answer 2

divide top and bottom by the highest power of x

Answer 3

isnt that only when the limit is going to infinity?

Answer 4

oh I'm sorry I misread :P

Answer 5

Try L'Hopital's Rule.

Answer 6

direct subtitute x=0

Answer 7

YOu will end up with \(-\dfrac{2}{6} = -\dfrac{1}{3}\)

Answer 8

@RadEn then he will have zero in denominator.

Answer 9

L' Hospital Rule is NOT valid here... It is valid only for 0/0 of infinity/infinity form...

Between, Direct substitution gives 4/0 which is UNDEFINED...

Answer 10

but numerator no zero :P

Answer 11

@saloniiigupta95 is correct

Answer 12

But zero in denominator, hence undefined. It doesn't work :P

Answer 13

the answer IS undefined

Answer 14

me too, @TuringTest :)

Answer 15

Forgot about 0/0 sorry

Answer 16

This is why I'm having trouble with iy

Answer 17

yes, sorry @RadEn , I was just referring to salon being right about why l'Hospital is not valid. I would give you a medal too if I could.

Answer 18

nopes, just kidding :)

Answer 19

:P
@Luigi0210 you can show that the left and right hand limits are different, so if you want to be thorough...

Answer 20

If you graph it, you can see that there is vertical asymptote at x = 0 and what it approaches from left side is different to what it approaches from right side so there is no limit at x= 0.

Answer 21

@TuringTest ... *saloni*... and if the matter is about the medal, you can UNDO the best response tab... No worries...

@Luigi0210 ... Do it like
RHL h==>0+ then replace x by +h... get a value...
LHL h==>0- then replace x by -h... You would be getting a different value
So, the Limit DOES NOT EXIST as far as I think...

Answer 22

\[\large \lim_{x \rightarrow 0}\dfrac{4-2x^3+x}{8x+6x^3} = \nexists\]

Answer 23

/thread

Answer 24

Split up the fraction. Each part except 4/(8x+6x^3) goes to an value. In this one you can see that when x is small and negative we head to negative infinity. If it's small and a positive we go to infinity.

Answer 25

Is it solved ??? @Luigi0210