Notice that
$$\lim_{x \to \infty} \sqrt{x^2-6x +1}-x = -3$$
$$\lim_{x \to -\infty} \sqrt{x^2-6x +1}-x = \infty$$
On manipulating the function by rationalization, I get:
$$\frac{-6+\frac1x}{\sqrt{1-\frac6x +\frac1{x^2}}+1}$$
What is the difference between $x$ approaching $\infty$ and x approaching $-\infty$? From what I see, all $x$ terms should approach $0$ regardless of whether $x$ approaches $\infty$ or approaches $-\infty$.
What is wrong with this thinking?
Best Answer
Remember that $\sqrt{x^2}=|x|$ (and not simply $x$). For $x\not=0$, $$\sqrt{x^2-6x +1}-x=\frac{(\sqrt{x^2-6x +1})^2-x^2}{\sqrt{x^2-6x +1}+x}=\frac{-6x +1}{\sqrt{x^2}\cdot \sqrt{1-6/x +1/x^2}+x}\\ =\frac{-6 +\frac{1}{x}}{\frac{|x|}{x}\cdot \sqrt{1-\frac6x +\frac1{x^2}}+1}.$$