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Prove statement is true

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove following statement is true or find a counterexample to demonstrate it is false?

    2. Relevant equations

    If lim x->3 f(x) / g(x) = 1 then lim x->3 (f(x)-g(x)) = 0

    3. The attempt at a solution

    well 3-3 = 0 and 3/3 = 1/1 = 1? But I think it's more than that
     
    Last edited: Jan 28, 2013
  2. jcsd
  3. Jan 28, 2013 #2

    jbunniii

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    Can you find an example where ##\lim_{x \rightarrow 3^+} \frac{f(x)}{g(x)} = 1## but ##\lim_{x \rightarrow 3^+} f(x)## and ##\lim_{x \rightarrow 3^+} g(x)## do not exist?
     
  4. Jan 28, 2013 #3
    sorry - I would like to fix the limit! I meant lim x->3 without the "+"
     
  5. Jan 28, 2013 #4

    Dick

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    jbunniii's hint doesn't really depend on the limit being one sided. Think about it.
     
  6. Jan 28, 2013 #5
    can i just pick any number or? because any same number divided by itself = 1 unless the answer wouldn't fit with the other functions
     
  7. Jan 28, 2013 #6

    Dick

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    You don't pick a number. You want to pick two functions of x. jbunniii's hint is that if lim x->3 of f(x) and g(x) both exist, you won't succeed. What an example of a function whose limit doesn't exist as x->3?
     
  8. Jan 28, 2013 #7

    Dick

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  9. Jan 28, 2013 #8
    lim x-> 3 (x^2-9)/(x-3)? Sorry if I'm way off or all over the place? I'm unsure.

    Edit:

    (x-3)/(x-3) from the first part.
     
  10. Jan 28, 2013 #9
  11. Jan 28, 2013 #10

    jbunniii

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    You won't be able to use exactly the functions from the first part, but putting (x-3) in the denominator is a good idea. See if you can find f and g such that [itex]\lim_{x \rightarrow 3} f(x) = \infty[/itex] and [itex]\lim_{x \rightarrow 3} g(x) = \infty[/itex] which satisfy [itex]\lim_{x \rightarrow 3} \frac{f(x)}{g(x)} = 1[/itex] but [itex]\lim_{x \rightarrow 3} f(x) - g(x) \neq 0[/itex]. (By choosing correctly, you can make that last limit equal to anything you want.)
     
  12. Jan 28, 2013 #11

    Dick

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    No, no fault here. But they are interrelated enough that you should be able to use one to help solve the other. What about trying to use f(x)=x/(x-3) and g(x)=3/(x-3) on this problem as well? What are the limits of f(x)/g(x) and f(x)-g(x) as x->3?
     
    Last edited: Jan 28, 2013
  13. Jan 28, 2013 #12
    I hate to say it. But, it's not clicking for me. So far I have x-3 in the denominator.

    I'm not too savvy with these functions and limits and am having trouble understanding the relations. I think I'm overlooking it most likely. These problems are more than we do in class purposely, so in class we don't do problems like these, so it's harder for me to comprehend what's going on when no examples like these have been taught. My apologies. It's quite frustrating.
     
  14. Jan 28, 2013 #13

    Dick

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    Hang in there, but didn't you already figure out lim x->3 of f(x)-g(x)=1 in the previous problem? That's half the battle. Remind me you how did it. Then use algebra to simplify f(x)/g(x) and try to find the limit of that. f(x)=x/(x-3) and g(x)=3/(x-3).
     
  15. Jan 28, 2013 #14
    I honestly had a friend do the f(x)-g(x) = 1

    But, I do understand from the first part that lim x->3+ (x/x-3) = 3/0+ = +∞
    and for g(x) (3)/(x-3) = 3/0+ = +∞
    but as for coming up with lim x->3+ (x-3)/(x-3) = 1 as a pair function with lim x->3+ (f(x)-g(x)) ≠0 I'm not sure.
     
  16. Jan 28, 2013 #15

    Dick

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    Well, that's honest. Is your algebra the problem? x/(x-3)-3/(x-3) should be easy to simplify. It's just like 1/3-2/3. Likewise, (x/(x-3))/(3/(x-3)) shouldn't be much of a challenge. It's just like (1/3)/(2/3). Can you do the number problems? Simplify the algebra problems with a variable the same way.
     
    Last edited: Jan 28, 2013
  17. Jan 28, 2013 #16
    isn't dividing fractions like the same as multiplying the inverse? So 1/3 / 2/3 = 3/6 = 1/2

    My algebra isn't bad per se, it's just my direction in math is bad as of now. I don't know where to go when done with the question. With that being said I haven't added and subtracted fractions with variables in a while - I just looked it up now as a refresher, not bad at all! I remember now.

    edit: I get how my friend got the answer now though! because the "-" carries over to the numerator which is 3 hence top and bottom are x-3 and I'm assuming two of the same variables equal out to 1
     
    Last edited: Jan 28, 2013
  18. Jan 28, 2013 #17

    Dick

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    Yes, not so hard once you think back. So what are the simplified forms of f(x)-g(x) and f(x)/g(x)?? The first step is pure algebra. You don't think about the limits until later.
     
  19. Jan 28, 2013 #18
    As for lim x->3 f(x)/g(x)

    (x)/(x-3) / (3)/(x-3)

    (x)/(x-3) * (x-3)/(3)

    = x/3
    x=3 ? if this means anything since f(x)/g(x) = 1 ?

    sorry I posted this before seeing your response.
     
  20. Jan 28, 2013 #19

    Dick

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    f(x)/g(x)=x/3, when x is not 3, sure. You've got the algebra part. Now you just have to figure out the limit x->3 x/3. You are writing '?' marks, but I'm not sure what the question is.
     
  21. Jan 28, 2013 #20
    Sorry my "?" is in regards to see if I'm on the right track.

    As the lim x->3 x/3 = 1 since I just plugged in 3 so 3/3 = 1
     
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