# limit involving logs

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• May 23rd 2007, 02:55 AM
DudenOxford
limit involving logs
Can someone please help me with this problem?

Prove the limit relation

lim_x-->0 log(x+1)/x = 1 by using the definition of the derivative

Thanks in advance
• May 23rd 2007, 04:03 AM
CaptainBlack
Quote:

Originally Posted by DudenOxford
Can someone please help me with this problem?

Prove the limit relation

lim_x-->0 log(x+1)/x = 1 by using the definition of the derivative

Thanks in advance

Not sure what "by using the definition of the derivative" means in this context, but L'Hopitals rule works here:

$
\lim_{x \to 0} \log(x+1)/x =$
${\lim_{x \to 0} d/dx (\log(x+1))}\over {\lim_{x \to 0}d/dx (x)}$ $= \lim_{x \to 0} 1/(1+x) = 1
$

RonL
• May 24th 2007, 05:31 PM
ThePerfectHacker
Quote:

Originally Posted by DudenOxford
Can someone please help me with this problem?

Prove the limit relation

lim_x-->0 log(x+1)/x = 1 by using the definition of the derivative

Thanks in advance

No need to use L'Hopital's rule here.

Consider the function $f(x) = \ln (1+x)$.
This function is differenciable at zero, and furthermore,
$f'(0)=1$.

Hence, by definition of derivative,
$\lim_{x\to 0} \frac{\ln(1+x) - \ln (1+0)}{x-0} = \lim_{x\to 0} \frac{\ln (1+x)}{x} = 1$