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