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

Printable View

- May 23rd 2007, 02:55 AMDudenOxfordlimit 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 AMCaptainBlack
Not sure what "by using the definition of the derivative" means in this context, but L'Hopitals rule works here:

$\displaystyle

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

$

RonL - May 24th 2007, 05:31 PMThePerfectHacker
No need to use L'Hopital's rule here.

Consider the function $\displaystyle f(x) = \ln (1+x)$.

This function is differenciable at zero, and furthermore,

$\displaystyle f'(0)=1$.

Hence, by definition of derivative,

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