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Math Help - 2 Questions on L-Hopitals Rule

  1. #1
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    2 Questions on L-Hopitals Rule

    I have one question that says

    lim x-> (3^x - 1) / (2^x - 1)

    I made that into

    lim x -> 0 (ln3 x 3^x) / (ln2 x 2^x)

    Is that correct? Where would I go from there?


    Also I have a question

    lim y->0 ((sqroot 5y+25) - 5) / y

    I am not sure how I would do this problem.

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by KarlosK View Post
    I have one question that says

    lim x-> (3^x - 1) / (2^x - 1)

    I made that into

    lim x -> 0 (ln3 x 3^x) / (ln2 x 2^x)

    Is that correct? Where would I go from there?
    Substitute x = 0?


    Also I have a question

    lim y->0 ((sqroot 5y+25) - 5) / y

    I am not sure how I would do this problem.

    Thanks for the help!
    Are you forced to use L'Hospital's rule??

    Without L'Hospital's you can just rationalize the numerator for a simple solution.
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  3. #3
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    Note that the second limit is the derivative of \sqrt{5x} at x=5.
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  4. #4
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    It says solve using l hopitals rule. That is what we are covering in class. Our teacher does not teach us very well though
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Substitute x = 0?




    Are you forced to use L'Hospital's rule??

    Without L'Hospital's you can just rationalize the numerator for a simple solution.
    If i substitute in X wont I just wind up with 0 over 0 again?
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  6. #6
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    He meant to substitute x=0 in the formula you got: \lim_{x\to 0} {\ln3 \cdot 3^x\over \ln2 \cdot 2^x}.
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  7. #7
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    Quote Originally Posted by maddas View Post
    He meant to substitute x=0 in the formula you got: \lim_{x\to 0} {\ln3 \cdot 3^x\over \ln2 \cdot 2^x}.

    Isint any # to the x in the case going to be 0. Which would make the bottom and top both equal to 0. Unless I don't know about some rule? ln3 x 3^0= 0?
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  8. #8
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    3^0=0? Are you quite sure? Doesn't that mean that 0 = 0\cdot 3^1 = 3^0\cdot3^1 = 3^{0+1} = 3^1 = 3?
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