How to solve this limit $\lim _{x\to 1}\left(\frac{\ln \left(\left|x-2\right|\right)}{\ln \left(x\right)}\right)$?

Question

I need to solve this limit please, $\lim _{x\to 1}\left(\frac{\ln \left(\left|x-2\right|\right)}{\ln \left(x\right)}\right)$ I tried various ways to solve it but the best one was simplifying it and I got in the end 1*1 * $\lim _{x\to 1}\left(\frac{\left(\left|x-2\right|-1\right)}{x-1}\right)$ now my problem is how to solve those kind of limit, when absolute values are in. I know that we need to separate it for example 1- and 1+, but that when the operation is equals to 0, here $\left|x-2\right|$ when x=1 is not equals to 0. So, we can't discuss the possibilities right? Thanks in advance P.S: L'Hopital rule is not allowed, please do not use it nor the graphing.

Answer 1

Put $x=1+z$ so you have to calculate $$\lim _{z\to 0}\frac{\ln (|z-1|)}{\ln (z+1)}=\lim _{z\to 0}\frac{\ln (|-(1-z)|)}{\ln (z+1)}=\lim _{z\to 0}\frac{\ln (1-z)}{\ln (z+1)}=$$.

Because of the form $\frac 00$ we apply l'Hôpital's Rule and get easily $$\lim _{z\to 0}\frac{\ln (1-z)}{\ln (z+1)}=-1$$.

Answer 2

I am not sure the following answer satisfies you, which made the process (unnecessarily) more complicated.

First the absolute sign can be safely removed when $x$ is sufficiently close to $1$. Then changing the variable $t = \log x$, then as $x \to 1$, $t \to 0$, and the expression to be evaluated can be written as \begin{align} & \frac{\log(2 - e^t)}{t} = \log(1 + 1 - e^t)^{1/t} = \log[1 + (1 - e^t)]^{\frac{1}{1 - e^t}\times \frac{1 - e^t}{t}} = \frac{1 - e^t}{t} \times\log[1 + (1 - e^t)]^{\frac{1}{1 - e^t}} \end{align} Clearly, $$\lim_{t \to 0}\frac{1 - e^t}{t} = - \lim_{t \to 0}\frac{e^t - e^0}{t - 0} = -\left.(e^t)'\right|_{t = 0} = -1.$$ And the second part tends to $1$ as $t \to 0$ in view of $1 - e^t \to 0$ and $\lim_{x \to 0}(1 + x)^{1/x} = e$. Hence the result is $-1$.

Answer 3

You should have $$ \lim_{t\to0}\frac{\log(1+t)}{t}=1\tag{*} $$ available. You can also note that, for $x$ sufficiently near to $1$, $|x-2|=2-x$. Thus your limit can be written $$ \lim_{x\to 1}\frac{\log(2-x)}{x-1}\frac{x-1}{\log x}= \lim_{x\to 1}\frac{\log(2-x)}{x-1} \cdot\lim_{x\to 1}\frac{x-1}{\log x} $$ provided both limits exist. But they do, because $$ \lim_{x\to 1}\frac{\log(2-x)}{x-1}= \lim_{x\to 1}-\frac{\log(1+(1-x))}{1-x}=-1 $$ and $$ \lim_{x\to1}\frac{x-1}{\log x}= \lim_{x\to1}\frac{x-1}{\log(1+(x-1))}=1 $$ both because of (*).

Alternatively, you can use the Taylor expansion, by doing the substitution $t=x-1$, so the limit becomes $$ \lim_{t\to0}\frac{\log(1-t)}{\log(1+t)}= \lim_{t\to0}\frac{-t+o(t)}{t+o(t)}=-1 $$