- #1

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## Homework Statement

limit as x->1/3 of (2-6x)^2/(3x-1)(9x^2-1)

## Homework Equations

## The Attempt at a Solution

substituting gets 0/0, tried expanding but it doesnt work either, dunno what else is there left to do...

- Thread starter blaze33
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- #1

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limit as x->1/3 of (2-6x)^2/(3x-1)(9x^2-1)

substituting gets 0/0, tried expanding but it doesnt work either, dunno what else is there left to do...

- #2

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can you use l'hopital's rule?

Edit: l'hopital's rule won't necessarily work for this one forget it

try simplifying the numerator and then factoring it again while recognizing there is a difference of squares in the denominator

my explanation might be vague but i don't want to give it away

Edit: l'hopital's rule won't necessarily work for this one forget it

try simplifying the numerator and then factoring it again while recognizing there is a difference of squares in the denominator

my explanation might be vague but i don't want to give it away

Last edited:

- #3

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factor it to (12x+4)(3x+1)

factor it again to 4(3x+1)(3x+1)

so

4(3x+1)(3x+1)/(3x-1)(9x^2-1)

can i cancel (3x+1) and (3x-1) ?

well i tried canceling it anyway and then substituted 1/3 again now i get 8/0 so its +infinity i guess.

- #4

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i took out the 4 from the beginning and got 4(9x^2 - 6x + 1) then got (3x-1)(3x-1)

you neglected the fact that it -24x, not +24x when you factored it

for the denominator: (3x-1)(9x^2 - 1)

recognize the difference of squares in the denominator?

it will become(3x-1)(3x-1)(3x+1), so cancel and evaluate

just curious do you know about l'hopital's rule?

if you do, try to evaluate the limit using that rule and you see that the answer is different

just another example of when l'hopital's rule fails if you're interested in knowing that

- #5

Mark44

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No, this should be 4(3x - 1)(3x -1).simplifying the numerator gives 36x^2-24x+4

factor it to (12x+4)(3x+1)

factor it again to 4(3x+1)(3x+1)

It would have been simpler to recognize that (2 - 6x)^2 = (6x - 2)^2 = 4(3x - 1)^2.

Can you cancel 3 + 1 and 3 - 1?so

4(3x+1)(3x+1)/(3x-1)(9x^2-1)

can i cancel (3x+1) and (3x-1) ?

well i tried canceling it anyway and then substituted 1/3 again now i get 8/0 so its +infinity i guess.

- #6

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yea your right... im not too good at algebra anyway so it takes time till i realize things like thatNo, this should be 4(3x - 1)(3x -1).

It would have been simpler to recognize that (2 - 6x)^2 = (6x - 2)^2 = 4(3x - 1)^2.

Can you cancel 3 + 1 and 3 - 1?

- #7

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yea i got 4(9x^2 - 6x + 1) the first time as well but didnt know what to do with it further.. im bad at factoring

i took out the 4 from the beginning and got 4(9x^2 - 6x + 1) then got (3x-1)(3x-1)

you neglected the fact that it -24x, not +24x when you factored it

for the denominator: (3x-1)(9x^2 - 1)

recognize the difference of squares in the denominator?

it will become(3x-1)(3x-1)(3x+1), so cancel and evaluate

just curious do you know about l'hopital's rule?

if you do, try to evaluate the limit using that rule and you see that the answer is different

just another example of when l'hopital's rule fails if you're interested in knowing that

i heard about l'hopital's rule but never learned how to use it.. my prof. didnt teach us cuz he said sometimes it doesnt give the right answer so why bother

- #8

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did you see my post above? kinda pointed that out and helped you out a little more

- #9

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yea man thanks!did you see my post above? kinda pointed that out and helped you out a little more

- #10

Mark44

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So what did you get as your final expression before you took the limit?

- #11

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Thanks a lot.

- #12

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yea i got it thanks!

So what did you get as your final expression before you took the limit?

i looked at the question and did it all over again and i certainly missed that i should have done 4(3x - 1)(3x -1) instead of 4(3x +1)(3x +1)

and for the final expression i got:

4(3x-1)(3x-1)/(3x-1)(3x-1)(3x+1)

which gives 2 after substituting.

- #13

Mark44

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