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Silja Group Title

limx-0: (ln(1+bx)-ln(1-bx))/x

  • one year ago
  • one year ago

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  1. Silja Group Title
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    The correct answer is 2b. I need the way to this:-)

    • one year ago
  2. vinnv226 Group Title
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    When we directly substitute 0 for x, we get 0/0. This means we can use l'Hôpital's rule, which says to find the derivative of the top and of the bottom, and then try direct substitution. The derivative of the bottom is simply 1. Try to find the derivative of the top piece on your own and let me know if you need more assistance with it.

    • one year ago
  3. hartnn Group Title
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    hello @Silja \(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \) first you can use the log property that \(\ln a-\ln b = \ln (a/b)\) have you done that already ?

    • one year ago
  4. Silja Group Title
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    Yes i Know it. But i don't find the solution.

    • one year ago
  5. hartnn Group Title
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    ok, if you have attempted, please show us your work, we'll spot the error in it, this approach will be faster than we giving you all the steps...

    • one year ago
  6. Silja Group Title
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    Thats very Bad because I write from the mobile phone. And I can't upload a picture here or?

    • one year ago
  7. hartnn Group Title
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    rather one simpler way is to separate the 2 limits , \(\dfrac{\ln (1+bx)}{x}-\dfrac{\ln (1-bx)}{x}\) and treat them individually is this what you have done ?

    • one year ago
  8. Silja Group Title
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    No I had another way but I get always the solution 0

    • one year ago
  9. hartnn Group Title
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    let me walk you through then do you know the standard limit \(\lim \limits_{x \rightarrow 0}\dfrac{\ln (1+x)}{x}=...?\)

    • one year ago
  10. hartnn Group Title
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    we will use this : \(\lim \limits_{x \rightarrow 0}\dfrac{\ln (1+x)}{x}=1\) so to bring your function in the form of (1+variable) / variable form i will multiply and divide by 'b' in the first term (notice that the variable is bx and not just 'x')\ \(b\dfrac{\ln(1+bx)}{bx}\) got this step ?

    • one year ago
  11. hartnn Group Title
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    similarly for the 2nd term, now the variable is -bx (notice!) \(\large (-b)\dfrac{\ln(1+(-bx))}{(-bx)}\)

    • one year ago
  12. hartnn Group Title
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    so, now we distribute the limits, b (and -b) can be take out of limit as they are constants \(\large b\lim \limits_{bx \rightarrow 0}\dfrac{\ln (1+bx)}{bx}-(-b)\lim \limits_{-bx \rightarrow 0}\dfrac{\ln (1+(-bx))}{(-bx)}\) so now using our standard formula, both the limit will equal \(\large 1.\) and we get, \(\large b(1)-(-b)(1)=b+b=2b\) ask if you have doubts in any step :)

    • one year ago
  13. Silja Group Title
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    wonderful thank you very much :-*

    • one year ago
  14. hartnn Group Title
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    welcome ^_^

    • one year ago
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