<s>[INST] Obliques asymptotes of a logarithmic and rational Function with absolute value. I'm trying to find the obliques asymptotes of the following function:
$$f(x)=|x-1| \log\Big(\sqrt{x^2 +3x+3}-x-1\Big)$$
Its domain should be this: $D(f)=(-\infty, +\infty)$;
I discussed the absolute values: $$|x-1|=\left\{ 
\begin{array}{c}
x-1 <=>x\geq1  \\ 
1-x <=>x\lt 1
\end{array}
\right. \ \text{and}\  \ |x|=\left\{ 
\begin{array}{c}
x <=>x\geq0  \\ 
-x <=>x\lt0
\end{array}
\right.; $$
When $f(x)$ approaches to the left border: 
$$\lim_{x\to\ - \ \infty} (1-x) \log\Bigg((-x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)=+\infty \ ;$$
When it goes to the right border:
$$\lim_{x\to\ + \ \infty} (x-1) \log\Big(\sqrt{x^2 +3x+3}-x-1\Big)=-\infty \ ;$$
Now I want to find obliques asymptotes.
$$m = \lim_{x\to\ - \ \infty} \frac{f(x)}{x}=\lim_{x\to\ - \ \infty}\frac1x(1-x) \log\Bigg((-x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)= - \infty \ ;$$
For $x\to-\infty$ I do not have oblique asymptote. So, I check for the right border:
$$m = \lim_{x\to\ + \ \infty} \frac{f(x)}{x}=\lim_{x\to\ + \ \infty}\frac1x(x-1) \log\Bigg((x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)= \frac1x x \Bigg(1-\frac1x\Bigg) \log\Bigg(\frac12\Bigg)= \log\Bigg(\frac12\Bigg) \ ;$$
Here comes my problem, I'm not able to find $q$, the "known term".
$$q = \lim_{x\to\ + \ \infty} f(x) - mx=\lim_{x\to\ + \ \infty}(x-1) \log\Bigg((x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg) -\log\Bigg(\frac12\Bigg)x = + \infty\cdot 0 \ ;$$
But this is undetermined form and I'm not able to solve it.
Could someone point me out where I'm doing wrong or just give me a hint to find the correct solution? Thank you. 
 [/INST] We have
\begin{align}
\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)-\log\Bigl(\frac{1}{2}\Bigr)&=
\log\Bigl(\frac{x+2}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)-\log\Bigl(\frac{1}{2}\Bigr)\\
&=\log\Bigl(\frac{2\,x+4}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+\frac{x+3-\sqrt{x^2 +3\,x+3}}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+\frac{3\,x+6}{\bigl(\sqrt{x^2 +3\,x+3}+x+1\bigr)\bigl(\sqrt{x^2 +3\,x+3}+x+3\bigr)}\Bigr)\\
&\sim\frac{3}{4\,x}\quad\text{as}\quad x\to+\infty.
\end{align}
Then
\begin{align}
f(x)-\log\Bigl(\frac{1}{2}\Bigr)\,x&=
x\Bigl(\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)-\log\Bigl(\frac{1}{2}\Bigr)\Bigr)-\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)\\&\sim\frac34+\log2.
\end{align}
</s>