Mixing
Limit of quotient
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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
$$ lim_x to infty fracsqrt4 x^2 - 4x+5 = lim_x to infty fracsqrt4 - frac4x^21 + frac5x = 2.$$
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.
limits
edited 1 hour ago
Leucippus
18.9k92769
asked 1 hour ago
nox15
334
add a comment |Â
up vote
4
down vote
favorite
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.
limits
edited 1 hour ago
Leucippus
18.9k92769
asked 1 hour ago
nox15
334
2
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
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
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.
limits
edited 1 hour ago
Leucippus
18.9k92769
asked 1 hour ago
nox15
334
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.
limits
limits
edited 1 hour ago
Leucippus
18.9k92769
asked 1 hour ago
nox15
334
edited 1 hour ago
Leucippus
18.9k92769
asked 1 hour ago
nox15
334
edited 1 hour ago
Leucippus
18.9k92769
edited 1 hour ago
Leucippus
18.9k92769
edited 1 hour ago
Leucippus
18.9k92769
18.9k92769
asked 1 hour ago
nox15
334
asked 1 hour ago
nox15
334
asked 1 hour ago
nox15
334
334
2
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
1 hour ago
add a comment |Â
2
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
1 hour ago
2
2
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
1 hour ago
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
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
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)$
answered 58 mins ago
Soroush khoubyarian
588212
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
28 mins ago
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
23 mins ago
 |Â
show 3 more comments
up vote
3
down vote
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.
answered 1 hour ago
Martin Argerami
117k1071165
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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)$
answered 58 mins ago
Soroush khoubyarian
588212
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
28 mins ago
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
23 mins ago
 |Â
show 3 more comments
up vote
1
down vote
accepted
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)$
answered 58 mins ago
Soroush khoubyarian
588212
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
28 mins ago
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
23 mins ago
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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)$
answered 58 mins ago
Soroush khoubyarian
588212
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)$
answered 58 mins ago
Soroush khoubyarian
588212
edited 16 mins ago
answered 58 mins ago
Soroush khoubyarian
588212
answered 58 mins ago
Soroush khoubyarian
588212
answered 58 mins ago
Soroush khoubyarian
588212
588212
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
28 mins ago
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
23 mins ago
 |Â
show 3 more comments
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
28 mins ago
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
23 mins ago
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
43 mins ago
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
43 mins ago
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.
– Soroush khoubyarian
40 mins ago
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.
– Soroush khoubyarian
40 mins ago
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
29 mins ago
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
29 mins ago
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
28 mins ago
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
28 mins ago
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
23 mins ago
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
23 mins ago
 |Â
show 3 more comments
up vote
3
down vote
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.
answered 1 hour ago
Martin Argerami
117k1071165
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
add a comment |Â
up vote
3
down vote
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.
answered 1 hour ago
Martin Argerami
117k1071165
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
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.
answered 1 hour ago
Martin Argerami
117k1071165
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.
answered 1 hour ago
Martin Argerami
117k1071165
edited 1 hour ago
answered 1 hour ago
Martin Argerami
117k1071165
answered 1 hour ago
Martin Argerami
117k1071165
answered 1 hour ago
Martin Argerami
117k1071165
117k1071165
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
add a comment |Â
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I don't quite understand. Could you elaborate on it?
– nox15
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
I'm sure you think that $sqrtx^2=x$. Try it with $x=-5$.
– Martin Argerami
1 hour ago
1
1
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 $x = - a$.
– Leucippus
1 hour ago
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 $x = - a$.
– Leucippus
1 hour ago
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
45 mins ago
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
45 mins ago
add a comment |Â
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2
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
1 hour ago