## 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

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

2. vinnv226 Group Title

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.

3. hartnn Group Title

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 ?

4. Silja Group Title

Yes i Know it. But i don't find the solution.

5. hartnn Group Title

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...

6. Silja Group Title

Thats very Bad because I write from the mobile phone. And I can't upload a picture here or?

7. hartnn Group Title

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 ?

8. Silja Group Title

No I had another way but I get always the solution 0

9. hartnn Group Title

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

10. hartnn Group Title

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 ?

11. hartnn Group Title

similarly for the 2nd term, now the variable is -bx (notice!) $$\large (-b)\dfrac{\ln(1+(-bx))}{(-bx)}$$

12. hartnn Group Title

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 :)

13. Silja Group Title

wonderful thank you very much :-*

14. hartnn Group Title

welcome ^_^