question on limits and their calculation In taking each of the limits
$$\lim_{x\to -\infty}\frac{x+2}{\sqrt {x^2-x+2}}\quad \text{ and } \quad \lim_{x\to \infty}\frac{x+2}{\sqrt {x^2-x+2}},$$
I find that both give the value $1$, although it should in fact be getting $-1$ and $1$, respectively.
This however doesn't show from the calculations...how does one solve this?
 A: You're right! First notice that
$$x^2-x+2\sim_\infty x^2$$
hence
$$\frac{x+2}{\sqrt{x^2-x+2}}\sim_\infty\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}$$
the result for $x\to+\infty$ is clear (and equals $1$) and for $-\infty$ we get
$$\lim_{x\to-\infty}\frac{x+2}{\sqrt{x^2-x+2}}=\lim_{x\to-\infty}\frac{x}{-x}=-1$$
A: It all boils down (simple algebra omitted, see other answers ) to showing what happens to 
$$
\lim_{x \to -\infty}\frac{x}{|x|}
$$
Remember the definition of the absolute value: for $x< 0 \ f(x) = -x$. Therefore your limit becomes $\lim_{x \to - \infty} \frac{x}{-x} = -1$. For $x \to \infty$ it is of course $1$.
A: $$=\frac{1+\frac{2}{x}}{\sqrt{1-\frac{1}{x}+\frac{2}{x^2}}}\rightarrow 1$$
(this is an answer to original question)
A: Your fundamental problem arises regarding the issue of signs when dealing with $x\to -\infty$ and it can be handled most easily (without applying too much thought and in almost mechanical fashion) by putting $x=-t$ and then letting $t \to\infty$. Thus we have $$\lim_{x\to -\infty}\frac{x+2}{\sqrt{x^{2}-x+2}}=\lim_{t \to \infty}\frac{-t+2}{\sqrt{t^{2} + t +2}}=-1$$ Note that this approach totally avoids the hassle of dealing with $|x|$ and the understanding that $\sqrt{x^{2}}=|x|$.
