Hello i have an assignment and this exercise and can't figure out on how to do this limit with taylor series, i know that by l'Hôpital rules the solution is 1, if i use taylor expansion the solution is 0, can anyone help me with this?
$$ \lim_{\ x \to 1} \frac{\ln(x)}{\ x-1} $$
One may set $$ u=x-1,\qquad x=1+u, $$ then as $x \to 1$ we have $u \to 0$, then by using a Taylor series expansion as $u \to 0$ one gets $$ \ln(1+u)=u+o(u) $$ or $$ \frac{\ln(1+u)}u=1+o(1) $$ yielding the wanted result.
$\ln$ is defferentiable in $1$ so, $$ \lim\limits_{h\to 0}\dfrac{\ln(1+h)-\overbrace{\ln(1)}^{=0}}{h}=\ln'(1)=1$$ So, we just need to set $h=x-1$ in the given limit to get this answer.
There is no need to complicate matters. Just recall the useful limit: $$\lim_{t \to 0} \frac{\ln(1+t)}{t} = 1$$
Now, just use the transformation $t = x - 1$ and see what happens.
with $f(x)=\ln(x)$ we have
$$f'(1)= \lim_{ x\to 1} \frac{f(x)-f(1)}{x-1}= \lim_{ x\to 1} \frac{\ln(x)-\ln(1)}{x-1}= \lim_{x\to1} \frac{\ln(x)}{x-1} = \frac{1}{x}\Bigg|_{x=1}=1$$
