Limit of quotient
Limit of quotient
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
$$ lim_x to infty fracsqrt4 x^2 - 4x+5 = lim_x to infty fracsqrt4 - frac4x^21 + frac5x = 2.$$
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
$$fracsqrt4-frac4x^21+frac5x,$$ 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.
3 Answers
3
As @Martin Argerami said, $sqrtx^2 = |x|$, and not $x$. When you want to solve a limit which has square roots, you have to do the following:
$lim_xto infty fracsqrt4x^2-4x+5=sqrtlim_xtoinftyfrac4x^2-4(x+5)^2$
The reason we could do this, is that when $xto infty$ 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 $xto -infty$, 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:
$lim_xto -infty fracsqrt4x^2-4x+5=-sqrtlim_xto-inftyfrac4x^2-4(x+5)^2$
$ lim sqrtf(x)^2 = pm lim f(x)$
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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?
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– nox15
Sep 15 '18 at 20:08
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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 $sqrtx^2$. 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.
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– Soroush khoubyarian
Sep 15 '18 at 20:11
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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.
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– nox15
Sep 15 '18 at 20:22
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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.
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– nox15
Sep 15 '18 at 20:23
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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)
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– Soroush khoubyarian
Sep 15 '18 at 20:28
You are making a mistake that is incredibly common, because it is taught very very poorly in school. The "secret" is that
$$
sqrtx^2=|x|,
$$
and not $x$. When $xgeq0$ you get $x$, but when $x<0$ you get $-x$.
In your manipulations, you wrote $$ fracsqrt4 x^2 - 4x+5 = fracsqrt4 - frac4x^21 + frac5x .$$ Try that "equality" with $x=-10$, for instance.
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I don't quite understand. Could you elaborate on it?
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– nox15
Sep 15 '18 at 19:42
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I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
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– 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 $sqrtx^2 = sqrta^2 = sqrt = |a|$, then signs must be considered. If $x geq 0$ then $x = +a$ and if $x leq 0$ then $x = - a$.
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– Leucippus
Sep 15 '18 at 19:46
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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?
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– nox15
Sep 15 '18 at 20:06
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You can; if you use that $sqrtx^2=|x|$. In your case, $sqrt4x^2-4=|x|,sqrt4-4/x^2$.
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– Martin Argerami
Sep 15 '18 at 22:11
$dfracsqrt4 cdot 5^25
= dfracsqrt4 cdot 5^2sqrt5^2
= sqrtdfrac4 cdot 5^25^2
= 2$
$dfracsqrt4 cdot (-5)^2-5
= dfracsqrt4 cdot (-5)^2-sqrt(-5)^2
= -sqrtdfrac4 cdot (-5)^2(-5)^2
= -2$
If $x < 0$, then $x = -sqrtx^2$.
$dfracsqrt4 cdot x^2x
= dfracsqrt4 cdot x^2-sqrtx^2
= -sqrtdfrac4 cdot x^2x^2
= -2$
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– Shaun
Sep 15 '18 at 19:48