limit involving logs

**DudenOxford** · May 23rd 2007, 02:55 AM

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

**CaptainBlack** · May 23rd 2007, 04:03 AM

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:

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

RonL

**ThePerfectHacker** · May 24th 2007, 05:31 PM

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

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