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$\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?

Salim
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    Pedantically speaking, L'Hospital's rule is circular in problems like these, because this limit is by definition $\left. \frac{d}{dx} \ln(x^2+1) \right |_{x=1}$. – Ian Jul 08 '18 at 17:56
  • How is that? x^2 + 1 - 2 = x^2 - 1 = (x-1)(x+1), which is not (x-1) which is the denominator. From what I know, I need (f(x) - f(a)) / (x-a) to conclude that it is a derivative. – Salim Jul 08 '18 at 18:12
  • $f(x)=\ln(x^2+1)$ in this situation. – Ian Jul 08 '18 at 18:15
  • Aah yes I see it now. Thank you very much! – Salim Jul 08 '18 at 18:16
  • @Ian I don't think that your comment was pedantic. – José Carlos Santos Jul 08 '18 at 18:34
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    @JoséCarlosSantos Some people disagree, because technically L'Hospital's rule gives the correct answer and thus is technically a valid method to solve a problem like this one. Thus going so far as to call it circular is perhaps a bit much, since you're just trying to do a calculation, not write a proof. But in my opinion, thinking about this kind of problem using L'Hospital's rule is mixing up concepts, which is why I say it this way. – Ian Jul 08 '18 at 18:44

5 Answers5

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Hint:

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

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

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

Surb
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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}

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