# 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.

• Really thank you! The result should be $\frac34 + \log(2)$. I think you have forgot to solve $\log\Big(\sqrt{x^2 +3x+3}-x-1\Big)$ but it should be fine. – NapMaster Jun 3 '18 at 11:31