Why is $\lim_{x \rightarrow -\infty}\frac{x}{\sqrt{x^2}} = -1$ and not $1$? I've been practicing horizontal asymptotes and I came across a problem that I do not understand.
I understood why $\lim_{x \rightarrow \infty}\frac{x - 2}{\sqrt{x^2  + 1}} = 1$, but i couldn't understand why $\lim_{x \rightarrow -\infty}\frac{x - 2}{\sqrt{x^2  + 1}} $ isn't $1$ as well, but $-1$ ?
When I calculated  $\lim_{x \rightarrow -\infty}\frac{x}{\sqrt{x^2}} $ in wolfram alpha, the result was $-1$.
If anyone could explain why $\lim_{x \rightarrow -\infty}\frac{x - 2}{\sqrt{x^2  + 1}} = -1$ or $\lim_{x \rightarrow -\infty}\frac{x}{\sqrt{x^2}} = -1$ 
I'd appreate it very much! 
 A: If $x<0, |x|=-x $ so ${x\over{\sqrt{x^2}}}$ is ${x\over{|x|}}={x\over{-x}}=-1$.
A: Note that
$$\sqrt{x^2} = \vert x\vert$$
so the following holds for negative values of $x$:
$$\sqrt{x^2} = -x; \quad x < 0$$
Hence, you get
$$\frac{x}{-x} = -1$$
A: Hint: Use that $$\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}$$
A: Hint:
$$
\lim_{x \rightarrow -\infty}\frac{x - 2}{\sqrt{x^2  + 1}} = 
\lim_{y \rightarrow \infty}\frac{-y - 2}{\sqrt{(-y)^2  + 1}} = -1
$$
A: The answer really has nothing to do with limits or asymptotes. The real square root function always returns the nonnegative square root. So for every negative value of $x$
$$
\frac{x}{\sqrt{x^2}} = \frac{x}{|x|} = -1 .
$$
A: The core issue is that $\sqrt{x^2}$ does not simplify to $x$ for negative $x$. 
The symbol $\sqrt{r}$ for a nonnegative real number $r$ denotes the nonnegative root of $r$. This is not altered by the fact that in this case $r=x^2$ for some (negative) $x$. 
Generally for a real number $x$ one has that $\sqrt{x^2} = |x|$. 
Keeping that in mind the issue should resolve itself. 
It's also that phenomenon for $\sqrt{x^2+1}$. When $x$ tends to $-\infty$ this is not close to $x$ but rather close to $|x|$.  
A: Because when $x\to\infty^{+}$, $x$ tends to numbers positives and when $x\to\infty^{-}$, $x$ tends to numbers negatives. If you develop this:
\begin{eqnarray}
\lim_{x\to\infty^{+}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{+}}\frac{x}{|x|}  \\
\lim_{x\to\infty^{+}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{+}}\frac{x}{x}  \\
\lim_{x\to\infty^{+}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{+}}1  \\
\lim_{x\to\infty^{+}}\frac{x}{\sqrt{x^2}} &=& 1  \\
\end{eqnarray}
And:
\begin{eqnarray}
\lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}\frac{x}{|x|} \\
\lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}\frac{x}{-x} \\
\lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}-1 \\
\lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& -1 \\
\end{eqnarray}
A: Set $y=-x$, where $y>0$(why?).
$-\dfrac{y}{\sqrt{y^2}} = - \dfrac{y}{|y|}=-\dfrac{y}{y}=-1.$
A: If you consider that $\frac{-2}{(x^2 +1)^{\frac{1}{2}}}$ goes to $0$, then let us just observe $\frac{x}{(x^2 +1)^{\frac{1}{2}}}$
$$= \frac{1}{\frac{1}{x} (x^2 +1)^{\frac{1}{2}}}$$
$$= \frac{1}{(1+ \frac{1}{x^2})^{\frac{1}{2}}}$$
Conceptually, is it now clear why that limit is $1$?
