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Silja

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

  • 9 months ago
  • 9 months ago

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

    • 9 months ago
  2. vinnv226
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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.

    • 9 months ago
  3. hartnn
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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 ?

    • 9 months ago
  4. Silja
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    Yes i Know it. But i don't find the solution.

    • 9 months ago
  5. hartnn
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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...

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

    • 9 months ago
  7. hartnn
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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 ?

    • 9 months ago
  8. Silja
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    No I had another way but I get always the solution 0

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

    • 9 months ago
  10. hartnn
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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 ?

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

    • 9 months ago
  12. hartnn
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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 :)

    • 9 months ago
  13. Silja
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    wonderful thank you very much :-*

    • 9 months ago
  14. hartnn
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    welcome ^_^

    • 9 months ago
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