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Homework Help: Limit for basic Calculus I

  1. Sep 29, 2013 #1
    lim of (3^y-5^y)/(2^y-7^y) as y->0

    Evaluate the above limit.

    Okay, so, I was flabbergasted at how challenging this problem is. I realized it was an indeterminate form [0/0]. I was trying to find a way to cancel y out of the numerator and denominator, in vain. I tried finding a pattern to expand the numerator and denominator, in vain: I still ended up with the [0/0].

    Any tips?

    Please, I have been racking my brains for 3.5 hours. :(
  2. jcsd
  3. Sep 29, 2013 #2


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    I would start by factoring a ##3^y## out of the numerator and a ##2^y## out of the denominator giving$$
    \left (\frac 3 2\right)^y \frac{1-(\frac 5 3)^y}{1 -(\frac 7 2)^y}$$and see if that gives you any ideas.
  4. Sep 29, 2013 #3
    I feel really lost and desperate :S.
    I tried taking the (5/3)^y and the (7/2)^y out again. No results. I'm trying to find a way of getting rid of that y.

    I have the answer in front of my face: ln5-ln3 over ln7-ln2
    I don't see how they get those logs in there :S

    Please be patient with my slow mind :P I have been studying from 8AM to 9PM.

    Last edited: Sep 29, 2013
  5. Sep 29, 2013 #4


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    Use your 'special limit'. Have you figured out what lim x->0 (a^x-1)/x is yet? It's 1 if a=e. What is it in general? That's a great place to start.
  6. Sep 29, 2013 #5
    Okay, I am lost.

    I made a mistake using the logarithm rules. I came to an answer that was close to the right answer, which excited me a bit. :(

    Dick: I just don't see how I could get y at the denominator to use that special limit.
  7. Sep 29, 2013 #6


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    Divide the numerator AND the denominator of LCKurtz's fraction by y. It doesn't change the result, right?
  8. Sep 29, 2013 #7
    Yes, I could find the answer to my previous question. Turns out I didn't need that 'special limit'. Indeed, I just multiplied the denominator and numerator by a^x and that did the trick in me elucidating that daunting question. Sorry I didn't have time to follow up! I have exams this week, so it's rush week!
  9. Sep 29, 2013 #8


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    It's fine if you don't have time to follow up. But what did you use instead of the 'special limit'? I don't think just multiplying by a^x will do it. If you think you've got a solution that way it's likely wrong. Just saying...
  10. Sep 29, 2013 #9
    Could you use L'Hopitals?
  11. Sep 29, 2013 #10
    Oh no, do not worry. I got the right answer. Actually, I may have used the 'special limit'. I don't recall. I will have to search through my pile of papers. I'll keep you posted on that.

    To answer johnqwertyful, no I cannot. I didn't learn it yet. That's what's making this problem tough :(
  12. Sep 29, 2013 #11
    I'm stuck.
    Same [0/0] story again.
  13. Sep 29, 2013 #12


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    Prove a useful result first. As I asked before, what is lim x->0 (a^x-1)/x??
  14. Sep 29, 2013 #13
    1 evidently! :)
  15. Sep 29, 2013 #14
    After 4 hours of sweat, I got it. :) You guys/girls are awesome. :)
  16. Sep 29, 2013 #15


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    That's not right. The limit is equal to 1 only if a=e.
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