Limit as x approaches 1/3: (2-6x)^2 / (3x-1)(9x^2-1) | Homework Help

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In summary, the limit as x approaches 1/3 of (2-6x)^2/(3x-1)(9x^2-1) can be evaluated by simplifying the numerator, factoring to recognize a difference of squares in the denominator, and then canceling like terms. Using l'hopital's rule is not necessary and may not give the correct answer.
  • #1
blaze33
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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 doesn't work either, don't know what else is there left to do...
 
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  • #2
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
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
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
blaze33 said:
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.
blaze33 said:
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
Mark44 said:
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... I am not too good at algebra anyway so it takes time till i realize things like that
 
  • #7
physicsman2 said:
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.. I am bad at factoring
i heard about l'hopital's rule but never learned how to use it.. my prof. didnt teach us because he said sometimes it doesn't give the right answer so why bother
 
  • #8
did you see my post above? kinda pointed that out and helped you out a little more
 
  • #9
physicsman2 said:
did you see my post above? kinda pointed that out and helped you out a little more

yea man thanks!
 
  • #10
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
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
Mark44 said:
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
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.
 

1. What is a limit in mathematics?

A limit is a fundamental concept in mathematics that describes the behavior of a function as the input values approach a certain value. It is used to determine the value that a function approaches but may never actually reach.

2. Why is finding a limit important?

Finding a limit is important because it allows us to understand the behavior of a function and make predictions about its values. It is also a crucial tool in calculus and other areas of mathematics.

3. How do I find a limit?

To find a limit, you can use various methods such as plugging in values, using algebraic manipulation, or using the properties of limits. It is important to understand the rules and techniques for evaluating limits, as well as knowing when certain methods are appropriate.

4. What are the common types of limits?

The most common types of limits are one-sided limits, which involve approaching a value from either the left or right side, and two-sided limits, which involve approaching a value from both sides. Other types include infinite limits, where the function approaches positive or negative infinity, and limits at infinity, where the function approaches a specific value as the input grows larger or smaller.

5. How can I use limits to solve real-world problems?

Limits can be used to solve real-world problems by modeling real-life situations with mathematical functions and determining the behavior of these functions using limits. This allows us to make predictions and optimize solutions in various fields such as physics, economics, and engineering.

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