# 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!

• $\sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative. – tilper Jan 2 at 15:12
• A plot on Wolfram Alpha demonstrates it quickly ... – Martin R Jan 2 at 15:16

If $$x<0, |x|=-x$$ so $${x\over{\sqrt{x^2}}}$$ is $${x\over{|x|}}={x\over{-x}}=-1$$.

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

$$\frac{x}{-x} = -1$$

Hint: Use that $$\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}$$

Hint: $$\lim_{x \rightarrow -\infty}\frac{x - 2}{\sqrt{x^2 + 1}} = \lim_{y \rightarrow \infty}\frac{-y - 2}{\sqrt{(-y)^2 + 1}} = -1$$

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 $$x$$ $$\frac{x}{\sqrt{x^2}} = \frac{x}{|x|} = -1 .$$

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 $$|x|$$.

Set $$y=-x$$, where $$y>0$$(why?).

$$-\dfrac{y}{\sqrt{y^2}} = - \dfrac{y}{|y|}=-\dfrac{y}{y}=-1.$$

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: $$\begin{eqnarray} \lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}\frac{x}{|x|} \\ \lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}\frac{x}{-x} \\ \lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& \lim_{x\to\infty^{-}}-1 \\ \lim_{x\to\infty^{-}}\frac{x}{\sqrt{x^2}} &=& -1 \\ \end{eqnarray}$$

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 $$1$$?

• Ah, as $x$ tends toward minus infinity, it is not not the same as the root of the square. I see what you're saying now. – Ryan Goulden Feb 3 at 5:08