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

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

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

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Yes i Know it. But i don't find the solution.
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...
Thats very Bad because I write from the mobile phone. And I can't upload a picture here or?
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 ?
No I had another way but I get always the solution 0
let me walk you through then do you know the standard limit \(\lim \limits_{x \rightarrow 0}\dfrac{\ln (1+x)}{x}=...?\)
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 ?
similarly for the 2nd term, now the variable is -bx (notice!) \(\large (-b)\dfrac{\ln(1+(-bx))}{(-bx)}\)
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 :)
wonderful thank you very much :-*
welcome ^_^

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