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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!

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    $\begingroup$ $\sqrt{x^2} = |x|$. when $x < 0$, $x / |x|$ is also negative. $\endgroup$ – tilper Jan 2 at 15:12
  • $\begingroup$ A plot on Wolfram Alpha demonstrates it quickly ... $\endgroup$ – Martin R Jan 2 at 15:16
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If $x<0, |x|=-x $ so ${x\over{\sqrt{x^2}}}$ is ${x\over{|x|}}={x\over{-x}}=-1$.

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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

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

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

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

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

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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 $|x|$.

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

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

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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}

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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 $1$?

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  • $\begingroup$ 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. $\endgroup$ – Ryan Goulden Feb 3 at 5:08

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