$\lim_{x \to 1} \dfrac{\ln(x^2+1)-\ln(2)}{x-1} $ = 1, why? $\lim_{x \to 1} \dfrac{\ln(x^2+1)-\ln(2)}{x-1} $ = 1
There is a same topic, but it did not help me to understand this problem. Could anyone shed some light on this?
I alreadty know that it is converging to 1, but how could I proof that, WITHOUT using l'Hopital rule?
 A: Hint:
$\dfrac{\ln(x^2+1)-\ln(2)}{x-1}$ is the rate of variation of the function $\ln(x^2+1)$ from $x=1$.
A: For a more general approach, that works even if the quotient cannot be easily identified with a derivative. We have 
$$
\log(x^2+1)=\log(2+(x^2-1))=\log(2(1+(x^2-1)/2))=\log 2+ \log (1+(x^2-1)/2).
$$
Using the approximation of $\log(1+t)$ at $t=0$, 
$$
\log(x^2+1)=\log 2+ \frac{x^2-1}2+o((x^2-1)^2). 
$$
Thus, with $x^2-1=(x-1)(x+1)$, 
$$
\dfrac{\ln(x^2+1)-\ln(2)}{x-1}
=\dfrac{\frac{x^2-1}2+o((x^2-1)^2)}{x-1}
={\frac{x+1}2}+o(x-1)\xrightarrow[x\to1]{}=\frac{1+1}2=1.
$$
A: By definition,
$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$
If $a=1$ and $f(x)=\ln(x^2+1)$ you have immediately 
$$\lim_{x\to 1}\frac{\ln(x^2+1)-\ln(2)}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=\left.\frac{d}{dx}\right|_{x=1}\ln(x^2+1)=\left. \frac{2x}{x^2+1}\right|_{x=1}=1.$$ 
A: By setting $y=x-1$, the limit becomes
\begin{align}
\lim_{y\to0}\frac{\log\left(1+y+\frac{1}{2}y^2\right)}{y}
    &=\lim_{y\to0}\frac{\log\left(1+y+\frac{1}{2}y^2\right)}{y+\frac{1}{2}y^2}\cdot\frac{y+\frac{1}{2}y^2}{y}=\\
    &=\lim_{y\to0}1\cdot\frac{1+\frac{1}{2}y}{1}=1
\end{align}
A: L'Hopital's Rule can be applied which tells you the limit. 
