I'm trying to find the 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 proceed studying the limit on the left border: $$\lim_{x\to\ - \ \infty} |x-1| \log\Big(\sqrt{x^2 +3x+3}-x-1\Big);$$ $$\lim_{x\to\ - \ \infty} |x-1| \log\Bigg(|x|\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg);$$ I discuss the absolute values: $$|x-1|=\left\{ \begin{array}{c} x-1 <=>x\geq1 \\ 1-x <=>x\lt 1 \end{array} \right. \ \text{and}\ \ |x|=\left\{ \begin{array}{c} x <=>x\geq0 \\ -x <=>x\lt0 \end{array} \right.; $$
So I obtain: $$\lim_{x\to\ - \ \infty} (1-x) \log\Bigg((-x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg);$$ $$\lim_{x\to\ - \ \infty} (1-x) \log\Bigg[(-x)\Bigg(\sqrt{1 +\frac3x+\frac3{x^2}}+1+\frac1x\Bigg)\Bigg];$$ $$\lim_{x\to\ - \ \infty} x\Bigg(-1+\frac1x\Bigg) \Bigg[\log(-x)+\log\Bigg(\sqrt{1 +\frac3x+\frac3{x^2}}+1+\frac1x\Bigg)\Bigg];$$ $$->+\infty\big[+\infty + \log(2)\big]=+\infty \ ;$$ While, on the right border: $$\lim_{x\to\ + \ \infty} (x-1) \log\Bigg(x\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg);$$ $$\lim_{x\to\ + \ \infty} x\Bigg(1-\frac1x\Bigg) \Bigg[\log(x)+\log\Bigg(\sqrt{1 +\frac3x+\frac3{x^2}}-1-\frac1x\Bigg)\Bigg];$$ $$->+\infty\big[+\infty + \log(0)\big]=+\infty \ ;$$ But, consulting Wolfram Alpha, the right limit should be: $-\infty$.
Since my calculations are wrong, I'm not able to find obliques asymptotes.
Could someone point me out where I'm doing wrong or just give me a hint to find the correct solution? Thank you.
Perhaps this will help: $$\sqrt{b}-\sqrt{a} = {b-a\over \sqrt{b}+\sqrt{a}}$$
so
$$f(x)=|x-1| \log\Big({x^2+3x+3-x^2-2x-1\over \sqrt{x^2 +3x+3}+x+1}\Big)$$ $$ = |x-1| \log\Big({x+2\over \sqrt{x^2 +3x+3}+x+1}\Big)$$
so $$ \lim _{x\to \infty}f(x)= \lim _{x\to \infty}|x-1| \log\underbrace{\Big({x(1+2/x)\over x(\sqrt{1 +3/x+3/x^2}+1+1/x)}\Big)}_{\longrightarrow {1\over 2}}= -\infty$$ since $\log (1/2)<0$