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I recently learned that when you are solving for the limit of a quotient, you have to divide everything by the highest number in the denominator, like this

$$ \lim_{x \to \infty} \frac{\sqrt{4 x^2 - 4}}{x+5} = \lim_{x \to \infty} \frac{\sqrt{4 - \frac{4}{x^2}}}{1 + \frac{5}{x}} = 2.$$

But I don't quite understand how when $x \to - \infty$ , the answer changes to $-2$, since when you divide everything by the highest power, it gives $$\frac{\sqrt{4-\frac{4}{x^2}}}{1+\frac{5}{x}},$$ meaning that even if you put negative infinity in the place of $x$, it still only gives $0$ and leaves $2$ as the final answer.

Why does the answer change to $-2$? I understand that it must be $-2$ when I look at the graph, I just don't understand the algebra part of it.

Leucippus
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nox15
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 15 '18 at 19:48

3 Answers3

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You are making a mistake that is incredibly common, because it is taught very very poorly in school. The "secret" is that $$ \sqrt{x^2}=|x|, $$ and not $x$. When $x\geq0$ you get $x$, but when $x<0$ you get $-x$.

In your manipulations, you wrote $$ \frac{\sqrt{4 x^2 - 4}}{x+5} = \frac{\sqrt{4 - \frac{4}{x^2}}}{1 + \frac{5}{x}} .$$ Try that "equality" with $x=-10$, for instance.

Martin Argerami
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  • I don't quite understand. Could you elaborate on it? – nox15 Sep 15 '18 at 19:42
  • I'm sure you think that $\sqrt{x^2}=x$. Try it with $x=-5$. – Martin Argerami Sep 15 '18 at 19:43
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    What is being said is, for example $x = \pm a$, leads to $x^2 = a^2$. When taking a square root, as in $\sqrt{x^2} = \sqrt{a^2} = \sqrt{|a|^2} = |a|$, then signs must be considered. If $x \geq 0$ then $x = +a$ and if $x \leq 0$ then $x = - a$. – Leucippus Sep 15 '18 at 19:46
  • I think I haven't made myself clear. I understand how you can derive -2 from the equation √(4x^2-4)/(x+5). I also know that √x^2=absolute value of x. What I don't get is how the method of dividing everything by the highest power and getting √(4-4/x^2)/(1+5/x) does not work with negative infinity. Even if I plug in negative infinity into √(4-4/x^2)/(1+5/x), it only gives 2 as the answer. Does that mean I can't use this method with limits in which x approaches negative infinity? – nox15 Sep 15 '18 at 20:06
  • You can; if you use that $\sqrt{x^2}=|x|$. In your case, $\sqrt{4x^2-4}=|x|,\sqrt{4-4/x^2}$. – Martin Argerami Sep 15 '18 at 22:11
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As @Martin Argerami said, $\sqrt{x^2} = |x|$, and not $x$. When you want to solve a limit which has square roots, you have to do the following:

$\lim_{x\to \infty} \frac{\sqrt{4x^2-4}}{x+5}=\sqrt{\lim_{x\to\infty}\frac{4x^2-4}{(x+5)^2}}$

The reason we could do this, is that when $x\to \infty$ the denominator is positive, the numerator is positive, hence the whole answer is positive. Our answer is under a square root which means our answer is consistent. (the square root of a real number is non negative.) However, when $x\to -\infty$, you can say that the result is negative, since the numerator is positive and the denominator is negative. In other words: $x + 5 = - \sqrt{(x + 5)^2}$, because $|x + 5| = -(x + 5)$. So: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-4}}{x+5}=-\sqrt{\lim_{x\to-\infty}\frac{4x^2-4}{(x+5)^2}}$

$ \lim \sqrt{f(x)^2} = \pm \lim f(x)$

Soroush khoubyarian
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  • Thank you. But I understand how you can derive -2 from the equation √(4x^2-4)/(x+5). I also know that √x^2=absolute value of x. What I don't get is how the method of dividing everything by the highest power and getting √(4-4/x^2)/(1+5/x) does not work with negative infinity. Even if I plug in negative infinity into √(4-4/x^2)/(1+5/x), it only gives 2 as the answer. Does that mean I can't use this method with limits in which x approaches negative infinity? – nox15 Sep 15 '18 at 20:08
  • It works, the only problem with square root is that the result is always positive. Essentially what I mean is that lets say you want to calculate $x$, and instead you calculate $\sqrt{x^2}$. The latter is always positive, regardless of the sign of $x$. When you want to take everything under the square root, you first have to find the sign of the answer(in this case negative since $x+5$ is negative), then add a negative sign behind the square root to make the answer negative. – Soroush khoubyarian Sep 15 '18 at 20:11
  • I perfectly understand what you mean. The numerator cannot be negative so it all depends on the denominator, and when the x is bigger than 5, the denominator becomes 5, thus giving a negative number and -2 as the limit when x is approaching negative infinity. I understand why it's -2. What I don't get is why the shortcut doesn't work with negative infinity. I learned that the shortcut is to divide everything by the highest power. So I do that. – nox15 Sep 15 '18 at 20:22
  • Then I get√(4-4/x^2)/(1+5/x) When x is approaching infinity, I just plug in infinity and see that it becomes √(4+0)/(1+0). So the limit is 2. I try that with the x is approaching negative infinity by plugging in negative infinity into the equation, which again gives √(4+0)/(1+0), or 2, instead of - 2. I just don't get why the shortcut wouldn't give me the right answer. – nox15 Sep 15 '18 at 20:23
  • The shortcut is ONLY AND ONLY true for rational functions. What you have is not rational, because the numerator is not a polynomial. What you have to do, is that you need to take the square root outside of the limit, then your limit ends up being a rational function. (Rational function is the division between two polynomials) – Soroush khoubyarian Sep 15 '18 at 20:28
  • Oh, now I get why it wasn't working! Thank you so much. But how can you take the square root outside of the limit? – nox15 Sep 15 '18 at 20:31
  • Look at my answer I've updated it – Soroush khoubyarian Sep 15 '18 at 20:33
  • It just shows a bunch of symbols. – nox15 Sep 15 '18 at 20:34
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$\dfrac{\sqrt{4 \cdot 5^2}}{5} = \dfrac{\sqrt{4 \cdot 5^2}}{\sqrt{5^2}} = \sqrt{\dfrac{4 \cdot 5^2}{5^2}} = 2$

$\dfrac{\sqrt{4 \cdot (-5)^2}}{-5} = \dfrac{\sqrt{4 \cdot (-5)^2}}{-\sqrt{(-5)^2}} = -\sqrt{\dfrac{4 \cdot (-5)^2}{(-5)^2}} = -2$

If $x < 0$, then $x = -\sqrt{x^2}$.

$\dfrac{\sqrt{4 \cdot x^2}}{x} = \dfrac{\sqrt{4 \cdot x^2}}{-\sqrt{x^2}} = -\sqrt{\dfrac{4 \cdot x^2}{x^2}} = -2$

Steven Alexis Gregory
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