How to get out of this indetermination $\lim\limits_{x\to -1} \frac{\ln(2+x)}{x+1}$? Well, is just that, I can't remember a way to get out of this indetermination (with logarithm), can someone help me?
I'm studying for my calculus test and this question is taking me some time.
$$\lim\limits_{x\to -1} \frac{\ln(2+x)}{x+1}$$
 A: $$
\lim_{x\to -1}\frac{\ln(2+x)}{x+1}=\lim_{x\to 0}\frac{\ln(1+x)}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)
$$
where $f(x)=\ln(1+x)$.
A: Hint:Take $w=(1+x)$ then you have the limit transformed as $\displaystyle\lim _{w\to 0}\frac{\ln (1+w)}{w}$
Solution :
$\displaystyle\ln(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}\dots=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{w^n}{n}$ ,for sufficiently small ${w}$
So we have ,
$\displaystyle\frac{\ln(1+w)}{w}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{w^{n-1}}{n}$
$\displaystyle\Rightarrow \lim_{w\to 0}\frac{\ln(1+w)}{w}=\lim_{w\to 0}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{w^{n-1}}{n}=1$ 
A: There are at least three methods:
$(1):$ Utilize $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$  putting $h=1+x$ (as I have suggested in the comment)
$(2):$ Use Maclaurin series of $\ln(1+y)$ by putting $y=1+x$
$(3):$ Use L'Hospital's Rule as $\lim_{x\to-1}\frac{\ln(2+x)}{x+1}$ is of the form $\frac00$
A: Since this is basically computing the derivative of the natural logarithm function, it's quite bizarre that using l'Hôpital's theorem is banned from the admitted tools.
If you're allowed to use substitutions, then you can change it into computing two limits.


*

*Change $1+x$ into $1/t$ (for $x>-1$), so you can write
$$\lim_{x\to -1^+}\frac{\ln(2+x)}{1+x}=
    \lim_{t\to\infty}\ln\left(1+\frac{1}{t}\right)^t=\ln e=1$$

*Change $1+x$ into $-1/t$ (for $x<-1$), so you can write
$$\lim_{x\to -1^-}\frac{\ln(2+x)}{1+x}=
    \lim_{t\to\infty}\ln\biggl(\biggl(1+\frac{-1}{t}\biggr)^t\,\biggr)^{-1}=\ln (e^{-1})^{-1}=1$$
Since both limits exist and are equal, you have your claim. However this depends on what ”fundamental limits” you're allowed to use.
