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Limit with L'Hopital's rule

  1. Dec 20, 2014 #1
    1. The problem statement, all variables and given/known data

    How can I fount the following limit using L'H'opetal rule?
    $$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$

    2. Relevant equations


    3. The attempt at a solution
    I tried to use the low
    $$\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a}e^{ln(f(x))}=e^l$$
    But it seems to be useless

    Many thanks for Help,.
     
  2. jcsd
  3. Dec 20, 2014 #2

    Dick

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    Start by using some rules of logs, like ln(a)-ln(b)=ln(a/b). Can you find the limit of the a/b part?
     
  4. Dec 20, 2014 #3
    $$\lim_{x\rightarrow+\infty}(ln(2x)-ln(x-4))$$
    $$\lim_{x\rightarrow+\infty}(ln \frac{2x}{x-4})$$

    Yes I can use L'H'opetal rule to find (a/b)

    $$\lim_{x\rightarrow+\infty}ln(\frac{2x}{x-4})$$
    $$\lim_{x\rightarrow+\infty}ln(\frac{2}{1})$$

    Oh,, it will be ln(2) .

    But how can I use L'H'opetal rule inside the (ln)?

    Thanks_For_Help
    :)
     
  5. Dec 20, 2014 #4

    Dick

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    Because ln is a continuous function. If ##\lim_{x\rightarrow+\infty} f(x)=L## and ##L>0## then ##\lim_{x\rightarrow+\infty} ln(f(x))=ln(L)##. Use l'Hopital to find the limit of the expression inside the ln.
     
  6. Dec 20, 2014 #5

    SteamKing

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    Why do you want to use l'Hopital's rule inside the log?
     
  7. Dec 20, 2014 #6
    I'm supposed to solve it by L'Hopital's rule .
    But I didn't think about another idea to solve it.
     
  8. Dec 20, 2014 #7
    Is there another way to do it?
     
  9. Dec 20, 2014 #8

    Mark44

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    See Dick's post #4.
     
  10. Dec 21, 2014 #9

    ehild

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    [tex]\lim_{x\rightarrow \infty} \frac{2x}{x-4}=\lim_{x\rightarrow \infty}\frac{2}{1-4/x}[/tex]
    4/x goes to zero ...
     
  11. Dec 21, 2014 #10
    The right answer is ln (2)

    Thus , I should evaluate the limit inside the (ln) then the final answer should be put in the (ln).
     
  12. Dec 21, 2014 #11

    ehild

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    Yes :)
     
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