# limit

• October 22nd 2009, 11:35 AM
the prince
limit
find the limit of the function:

lim {ln(1+x)/(3^x -1)}
x-> 0
• October 22nd 2009, 11:48 AM
ramiee2010
Quote:

Originally Posted by the prince
find the limit of the function:

lim {ln(1+x)/(3^x -1)}
x-> 0

$\lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}$
$= \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}$
• October 22nd 2009, 12:22 PM
the prince
why its $\lim_{x \to 0} \frac{ \ln (1+x)}{x} = 1$ and how did you find $\lim_{x \to 0} \frac{(3^x -1)}{x}= \log_{e}{3}$
(Thinking)
• October 22nd 2009, 12:30 PM
CaptainBlack
Quote:

Originally Posted by ramiee2010
$\lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}$
$= \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}$

This is useless without an explanation of what you think you are doing.

Also

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

needs to be justified as does:

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

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
• October 22nd 2009, 01:12 PM
ramiee2010
Quote:

Originally Posted by CaptainBlack
This is useless without an explanation of what you think you are doing.

Also

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

needs to be justified as does:

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

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

CB

these are identities which are taught us in 12th standard
$\lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right )$
$=\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right ) =1$

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

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

And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.
O.K.. Captain
• October 22nd 2009, 09:55 PM
CaptainBlack
Quote:

Originally Posted by ramiee2010
these are identities which are taught us in 12th standard
$\lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right )$
$=\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right ) =1$

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

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

Well they are not universally taught (because its a waste of brain space to remember things that can be rederived in 10 seconds) and so they need to be justified.

CB