find the limit of the function:

lim {ln(1+x)/(3^x -1)}

x-> 0

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- Oct 22nd 2009, 11:35 AMthe princelimit
find the limit of the function:

lim {ln(1+x)/(3^x -1)}

x-> 0 - Oct 22nd 2009, 11:48 AMramiee2010
$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}$

$\displaystyle = \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}$ - Oct 22nd 2009, 12:22 PMthe prince
why its $\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{x} = 1$ and how did you find $\displaystyle \lim_{x \to 0} \frac{(3^x -1)}{x}= \log_{e}{3}$

(Thinking)

can you give me a little advice please? - Oct 22nd 2009, 12:30 PMCaptainBlack
This is useless without an explanation of what you think you are doing.

Also

$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{x}=1$

needs to be justified as does:

$\displaystyle \lim_{x \to 0} \frac{(3^x -1)}{x}=\ln(3)$

Hence you may as well just have suggested the OP goes straight to L'Hopital's rule to find the limit.

And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.

CB - Oct 22nd 2009, 01:12 PMramiee2010
these are identities which are taught us in 12th standard

$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right ) $

$\displaystyle =\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right ) =1$

and

$\displaystyle \lim_{x \to 0} \frac{(a^x -1)}{x} = \lim_{x \to 0} \left ( \frac {1+x log_{e}{a}+ \frac {x^2}{2!} (log_{e}{a})^2+.... -1}{x} \right )$

$\displaystyle = \lim_{x \to 0} \left ( log_{e}{a}+ \frac {x}{2!} (log_{e}{a})^2+.... \right ) = log_{e}{a} $

Quote:

And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.

- Oct 22nd 2009, 09:55 PMCaptainBlack