Finding a limit help.

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  • #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...
 

Answers and Replies

  • #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
 
Last edited:
  • #3
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simplifying the numerator gives 36x^2-24x+4
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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actually i factored out the numerator a different way
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:cool:

just another example of when l'hopital's rule fails if you're interested in knowing that
 
  • #5
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simplifying the numerator gives 36x^2-24x+4
factor it to (12x+4)(3x+1)
factor it again to 4(3x+1)(3x+1)
No, 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.
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.
Can you cancel 3 + 1 and 3 - 1?
 
  • #6
14
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No, 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?
yea your right... im not too good at algebra anyway so it takes time till i realize things like that
 
  • #7
14
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actually i factored out the numerator a different way
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:cool:

just another example of when l'hopital's rule fails if you're interested in knowing that
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 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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did you see my post above? kinda pointed that out and helped you out a little more
yea man thanks!
 
  • #10
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I hope you realized that when I asked "can you cancel 3 + 1 and 3 - 1" you understood me to mean that you can't do that. And you can't cancel 3x + 1 and 3x - 1.

So what did you get as your final expression before you took the limit?
 
  • #11
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hey mark44, i was wondering if you could evaluate the limit using l'hopital's rule and see if I was right or not about it not working for this limit it'll probably only take you 5 seconds

Thanks a lot.
 
  • #12
14
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I hope you realized that when I asked "can you cancel 3 + 1 and 3 - 1" you understood me to mean that you can't do that. And you can't cancel 3x + 1 and 3x - 1.

So what did you get as your final expression before you took the limit?
yea i got it thanks!

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
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5,636
That's the right answer, but you can't just substitute 1/3 in because the numerator will be 0 and so will the denominator. That's where the limit comes in. As long as x is only close to 1/3, but not equal to it, then (3x - 1)^2 over itself will be close to 1, and the rest will be close to 4/2 = 2.
 

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