$\lim_{x \to 1} \dfrac{\ln(x^2+1)-\ln(2)}{x-1} $ = 1, why?

Question

$\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?

Answer 1

Hint:

$\dfrac{\ln(x^2+1)-\ln(2)}{x-1}$ is the rate of variation of the function $\ln(x^2+1)$ from $x=1$.

Answer 2

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 $x^2-1=(x-1)(x+1)$, $$ \dfrac{\ln(x^2+1)-\ln(2)}{x-1} =\dfrac{\frac{x^2-1}2+o((x^2-1)^2)}{x-1} ={\frac{x+1}2}+o(x-1)\xrightarrow[x\to1]{}=\frac{1+1}2=1. $$

Answer 3

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 $$\lim_{x\to 1}\frac{\ln(x^2+1)-\ln(2)}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=\left.\frac{d}{dx}\right|_{x=1}\ln(x^2+1)=\left. \frac{2x}{x^2+1}\right|_{x=1}=1.$$

Answer 4

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}

Answer 5

L'Hopital's Rule can be applied which tells you the limit.