Mixing
Why do the “rules” of horizontal asymptotes of rational functions work?
$begingroup$
Note: My current understanding is only at a college algebra level
From what I've seen online, in layman terms, the rules for horizontal asymptotes are as follows:
Rule 1) If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote will be $y=0$
Rule 2) If the numerator and denominator have equal degrees, then the horizontal asymptote will be a ratio of their leading coefficients
Rule 3) If the degree of the numerator is exactly one more than the degree of the denominator, then the oblique asymptote is found by dividing the numerator by the denominator. The resulting quotient is a linear expression which defines the oblique asymptote, and if there's a remainder, it's discarded.
For the first rule, I somewhat understand why the horizontal asymptote would be $y=0$
If the degree of the denominator is larger than the degree of the numerator, then the denominator is increasing at a faster rate than the numerator as $xrightarrowinfty$. The numerator "can't keep up" and it would be getting divided by increasingly larger values so the outputs would be getting smaller and smaller approaching $0$.
Am I on the right track with my thinking here?
For rule 2 I'm not sure why the ratio of the leading coefficients of the numerator and denominator are used as the horizontal asymptote.
For Rule 3 why divide the numerator by the denominator to get an oblique asymptote? Why isn't there a horizontal asymptote instead?
algebra-precalculus rational-functions
asked Jan 15 at 21:31
SleckerSlecker
1569
$endgroup$
add a comment |
$begingroup$
Note: My current understanding is only at a college algebra level
From what I've seen online, in layman terms, the rules for horizontal asymptotes are as follows:
Rule 1) If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote will be $y=0$
Rule 2) If the numerator and denominator have equal degrees, then the horizontal asymptote will be a ratio of their leading coefficients
Rule 3) If the degree of the numerator is exactly one more than the degree of the denominator, then the oblique asymptote is found by dividing the numerator by the denominator. The resulting quotient is a linear expression which defines the oblique asymptote, and if there's a remainder, it's discarded.
For the first rule, I somewhat understand why the horizontal asymptote would be $y=0$
If the degree of the denominator is larger than the degree of the numerator, then the denominator is increasing at a faster rate than the numerator as $xrightarrowinfty$. The numerator "can't keep up" and it would be getting divided by increasingly larger values so the outputs would be getting smaller and smaller approaching $0$.
Am I on the right track with my thinking here?
For rule 2 I'm not sure why the ratio of the leading coefficients of the numerator and denominator are used as the horizontal asymptote.
For Rule 3 why divide the numerator by the denominator to get an oblique asymptote? Why isn't there a horizontal asymptote instead?
algebra-precalculus rational-functions
asked Jan 15 at 21:31
SleckerSlecker
1569
$endgroup$
add a comment |
$begingroup$
Note: My current understanding is only at a college algebra level
From what I've seen online, in layman terms, the rules for horizontal asymptotes are as follows:
Rule 1) If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote will be $y=0$
Rule 2) If the numerator and denominator have equal degrees, then the horizontal asymptote will be a ratio of their leading coefficients
Rule 3) If the degree of the numerator is exactly one more than the degree of the denominator, then the oblique asymptote is found by dividing the numerator by the denominator. The resulting quotient is a linear expression which defines the oblique asymptote, and if there's a remainder, it's discarded.
For the first rule, I somewhat understand why the horizontal asymptote would be $y=0$
If the degree of the denominator is larger than the degree of the numerator, then the denominator is increasing at a faster rate than the numerator as $xrightarrowinfty$. The numerator "can't keep up" and it would be getting divided by increasingly larger values so the outputs would be getting smaller and smaller approaching $0$.
Am I on the right track with my thinking here?
For rule 2 I'm not sure why the ratio of the leading coefficients of the numerator and denominator are used as the horizontal asymptote.
For Rule 3 why divide the numerator by the denominator to get an oblique asymptote? Why isn't there a horizontal asymptote instead?
algebra-precalculus rational-functions
asked Jan 15 at 21:31
SleckerSlecker
1569
$endgroup$
Note: My current understanding is only at a college algebra level
From what I've seen online, in layman terms, the rules for horizontal asymptotes are as follows:
Rule 1) If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote will be $y=0$
Rule 2) If the numerator and denominator have equal degrees, then the horizontal asymptote will be a ratio of their leading coefficients
Rule 3) If the degree of the numerator is exactly one more than the degree of the denominator, then the oblique asymptote is found by dividing the numerator by the denominator. The resulting quotient is a linear expression which defines the oblique asymptote, and if there's a remainder, it's discarded.
For the first rule, I somewhat understand why the horizontal asymptote would be $y=0$
If the degree of the denominator is larger than the degree of the numerator, then the denominator is increasing at a faster rate than the numerator as $xrightarrowinfty$. The numerator "can't keep up" and it would be getting divided by increasingly larger values so the outputs would be getting smaller and smaller approaching $0$.
Am I on the right track with my thinking here?
For rule 2 I'm not sure why the ratio of the leading coefficients of the numerator and denominator are used as the horizontal asymptote.
For Rule 3 why divide the numerator by the denominator to get an oblique asymptote? Why isn't there a horizontal asymptote instead?
algebra-precalculus rational-functions
algebra-precalculus rational-functions
asked Jan 15 at 21:31
SleckerSlecker
1569
asked Jan 15 at 21:31
SleckerSlecker
1569
edited Jan 15 at 22:16
Slecker
asked Jan 15 at 21:31
SleckerSlecker
1569
asked Jan 15 at 21:31
SleckerSlecker
1569
asked Jan 15 at 21:31
SleckerSlecker
1569
1569
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $dfrac{c}{x} to 0$ as $x to infty$.
For the first case, you have the right idea. For instance, let $f(x) = dfrac{x^2+5x+100}{x^3-5}$. You need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty}frac{x^2+5x+100}{x^3-5} = lim_{x to infty}frac{x^2left(frac{5}{x}+frac{100}{x^2}right)}{x^3left(1-frac{5}{x^3}right)} = lim_{x to infty}frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x to infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $pm infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty} frac{ax^2+bx+c}{dx^2+ex+f} = lim_{x to infty} frac{x^2left(a+frac{b}{x}+frac{c}{x^2}right)}{x^2left(d+frac{e}{x}+frac{f}{x^2}right)} = frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x to infty$, so all that “remains” is $dfrac{ax^2}{dx^2} = dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x to infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$lim_{x to infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x to infty$, which will give $b$:
$$implies lim_{x to infty} frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = lim_{x to infty} frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$lim_{x to infty} frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
answered Jan 15 at 21:51
KM101KM101
6,0251525
$endgroup$
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
add a comment |
$begingroup$
You are on the right track in the first case. Similar considerations can apply to the other two cases you examine.
If the degrees are equal, then both are growing at the same rate. In that case, factor out of each the highest power: e.g. if the numerator is $P(x) = ax^n + p(x)$ where $p(x)$ is of lower degree, then write it as $P(x) = ax^n (1 + frac{p(x)}{ax^n})$. Now, when $x$ goes to $infty$, the second term goes to $0$. Similarly, the denominator $Q(x) = bx^n + q(x)$ where $q(x)$ is of lower degree, can be written as $Q(x) = bx^n(1 + frac{q(x)}{bx^n})$ and the second term of this also goes to $0$ as $x$ goes to $infty$. You are left with $frac{P(x)}{Q(x)} = frac{ax^n}{bx^n} = frac{a}{b}$.
For the third case, remember that the polynomial division algorithm allows you to write
$$ P(x) = Q(x)(ax + b) + R(x) $$
where the quotient is of degree 1 since $Q(x)$ has degree one less than $P(x)$, and
where the remainder $R(x)$ is of lesser degree than $Q(x)$. Dividing through by $Q(x)$ gives
$$ frac{P(x)}{Q(x)} = ax + b + frac{R(x)}{Q(x)}$$
Now if you let $x$ go to $infty$, the last term goes to 0, since the degree of $R(x)$ is less than the degree of $Q(x)$. IOW the LHS goes to $infty$ linearly i.e. just like $ax + b$ goes to $infty$. But the latter is a straight line with slope $a$.
That's the basic idea although you'll have to work a bit harder to "dot every i and cross every t."
answered Jan 15 at 21:55
NickDNickD
1,1211512
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2 Answers
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2 Answers
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$begingroup$
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $dfrac{c}{x} to 0$ as $x to infty$.
For the first case, you have the right idea. For instance, let $f(x) = dfrac{x^2+5x+100}{x^3-5}$. You need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty}frac{x^2+5x+100}{x^3-5} = lim_{x to infty}frac{x^2left(frac{5}{x}+frac{100}{x^2}right)}{x^3left(1-frac{5}{x^3}right)} = lim_{x to infty}frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x to infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $pm infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty} frac{ax^2+bx+c}{dx^2+ex+f} = lim_{x to infty} frac{x^2left(a+frac{b}{x}+frac{c}{x^2}right)}{x^2left(d+frac{e}{x}+frac{f}{x^2}right)} = frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x to infty$, so all that “remains” is $dfrac{ax^2}{dx^2} = dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x to infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$lim_{x to infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x to infty$, which will give $b$:
$$implies lim_{x to infty} frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = lim_{x to infty} frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$lim_{x to infty} frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
answered Jan 15 at 21:51
KM101KM101
6,0251525
$endgroup$
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
add a comment |
$begingroup$
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $dfrac{c}{x} to 0$ as $x to infty$.
For the first case, you have the right idea. For instance, let $f(x) = dfrac{x^2+5x+100}{x^3-5}$. You need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty}frac{x^2+5x+100}{x^3-5} = lim_{x to infty}frac{x^2left(frac{5}{x}+frac{100}{x^2}right)}{x^3left(1-frac{5}{x^3}right)} = lim_{x to infty}frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x to infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $pm infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty} frac{ax^2+bx+c}{dx^2+ex+f} = lim_{x to infty} frac{x^2left(a+frac{b}{x}+frac{c}{x^2}right)}{x^2left(d+frac{e}{x}+frac{f}{x^2}right)} = frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x to infty$, so all that “remains” is $dfrac{ax^2}{dx^2} = dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x to infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$lim_{x to infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x to infty$, which will give $b$:
$$implies lim_{x to infty} frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = lim_{x to infty} frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$lim_{x to infty} frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
answered Jan 15 at 21:51
KM101KM101
6,0251525
$endgroup$
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
add a comment |
$begingroup$
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $dfrac{c}{x} to 0$ as $x to infty$.
For the first case, you have the right idea. For instance, let $f(x) = dfrac{x^2+5x+100}{x^3-5}$. You need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty}frac{x^2+5x+100}{x^3-5} = lim_{x to infty}frac{x^2left(frac{5}{x}+frac{100}{x^2}right)}{x^3left(1-frac{5}{x^3}right)} = lim_{x to infty}frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x to infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $pm infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty} frac{ax^2+bx+c}{dx^2+ex+f} = lim_{x to infty} frac{x^2left(a+frac{b}{x}+frac{c}{x^2}right)}{x^2left(d+frac{e}{x}+frac{f}{x^2}right)} = frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x to infty$, so all that “remains” is $dfrac{ax^2}{dx^2} = dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x to infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$lim_{x to infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x to infty$, which will give $b$:
$$implies lim_{x to infty} frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = lim_{x to infty} frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$lim_{x to infty} frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
answered Jan 15 at 21:51
KM101KM101
6,0251525
$endgroup$
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $dfrac{c}{x} to 0$ as $x to infty$.
For the first case, you have the right idea. For instance, let $f(x) = dfrac{x^2+5x+100}{x^3-5}$. You need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty}frac{x^2+5x+100}{x^3-5} = lim_{x to infty}frac{x^2left(frac{5}{x}+frac{100}{x^2}right)}{x^3left(1-frac{5}{x^3}right)} = lim_{x to infty}frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x to infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $pm infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $lim_limits{x to infty} f(x)$:
$$lim_{x to infty} frac{ax^2+bx+c}{dx^2+ex+f} = lim_{x to infty} frac{x^2left(a+frac{b}{x}+frac{c}{x^2}right)}{x^2left(d+frac{e}{x}+frac{f}{x^2}right)} = frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x to infty$, so all that “remains” is $dfrac{ax^2}{dx^2} = dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x to infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$lim_{x to infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x to infty$, which will give $b$:
$$implies lim_{x to infty} frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = lim_{x to infty} frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$lim_{x to infty} frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
answered Jan 15 at 21:51
KM101KM101
6,0251525
edited Jan 16 at 6:37
answered Jan 15 at 21:51
KM101KM101
6,0251525
answered Jan 15 at 21:51
KM101KM101
6,0251525
answered Jan 15 at 21:51
KM101KM101
6,0251525
6,0251525
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
add a comment |
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Apologies in advance for splitting this reply up. Part (1) I used your same strategy for an arbitrary rational function of: $f(x)=frac{3x^3+6x^2+4x+2}{x^2+5x+2}$. First I factored out the largest degree from each polynomial $frac{x^3(3+frac{6x^2}{x^3}+frac{4x}{x^3}+frac{2}{x^3})} {x^2(1+frac{5x}{x^2}+frac{2}{x^2})}$, which simplifes to $frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$. I then used a limit to show that the terms with smaller degrees in the numerator than the denominator will become $0$ as $x$ goes to infinity.
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
$begingroup$
Part (2)$lim_{xto infty} frac{x^3(3+frac{6}{x}+frac{4}{x^2}+frac{2}{x^3})} {x^2(1+frac{5}{x}+frac{2}{x^2})}$ $=lim_{xto infty} frac{3x^3} {x^2}$ Which of course simplifies to $3x$, which is the slope of the oblique asymptote as $xrightarrowinfty$. BUT When I use polynomial division to divide the numerator by the denominator I get $3x-9$ (discarding the remainder), which is the actual equation of the oblique asymptote. Am I missing a step somewhere in my "argument" for the third case as to why the $-9$ wasn't included when using limits? Am I on the right track?
$endgroup$
– Slecker
Jan 16 at 4:50
1
1
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
That’s right. I was trying to give a simple explanation of why this happens (why it’s linearly slanted). The limit gives only the slope. In order to find the full evaluation with the constant term, you can use polynomial long division. I will add another explanation to the answer considering this.
$endgroup$
– KM101
Jan 16 at 6:26
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
$begingroup$
Thank you and cheers!
$endgroup$
– Slecker
Jan 16 at 20:38
1
1
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
$begingroup$
No problem, glad I was able to help!
$endgroup$
– KM101
Jan 16 at 20:45
add a comment |
$begingroup$
You are on the right track in the first case. Similar considerations can apply to the other two cases you examine.
If the degrees are equal, then both are growing at the same rate. In that case, factor out of each the highest power: e.g. if the numerator is $P(x) = ax^n + p(x)$ where $p(x)$ is of lower degree, then write it as $P(x) = ax^n (1 + frac{p(x)}{ax^n})$. Now, when $x$ goes to $infty$, the second term goes to $0$. Similarly, the denominator $Q(x) = bx^n + q(x)$ where $q(x)$ is of lower degree, can be written as $Q(x) = bx^n(1 + frac{q(x)}{bx^n})$ and the second term of this also goes to $0$ as $x$ goes to $infty$. You are left with $frac{P(x)}{Q(x)} = frac{ax^n}{bx^n} = frac{a}{b}$.
For the third case, remember that the polynomial division algorithm allows you to write
$$ P(x) = Q(x)(ax + b) + R(x) $$
where the quotient is of degree 1 since $Q(x)$ has degree one less than $P(x)$, and
where the remainder $R(x)$ is of lesser degree than $Q(x)$. Dividing through by $Q(x)$ gives
$$ frac{P(x)}{Q(x)} = ax + b + frac{R(x)}{Q(x)}$$
Now if you let $x$ go to $infty$, the last term goes to 0, since the degree of $R(x)$ is less than the degree of $Q(x)$. IOW the LHS goes to $infty$ linearly i.e. just like $ax + b$ goes to $infty$. But the latter is a straight line with slope $a$.
That's the basic idea although you'll have to work a bit harder to "dot every i and cross every t."
answered Jan 15 at 21:55
NickDNickD
1,1211512
$endgroup$
add a comment |
$begingroup$
You are on the right track in the first case. Similar considerations can apply to the other two cases you examine.
If the degrees are equal, then both are growing at the same rate. In that case, factor out of each the highest power: e.g. if the numerator is $P(x) = ax^n + p(x)$ where $p(x)$ is of lower degree, then write it as $P(x) = ax^n (1 + frac{p(x)}{ax^n})$. Now, when $x$ goes to $infty$, the second term goes to $0$. Similarly, the denominator $Q(x) = bx^n + q(x)$ where $q(x)$ is of lower degree, can be written as $Q(x) = bx^n(1 + frac{q(x)}{bx^n})$ and the second term of this also goes to $0$ as $x$ goes to $infty$. You are left with $frac{P(x)}{Q(x)} = frac{ax^n}{bx^n} = frac{a}{b}$.
For the third case, remember that the polynomial division algorithm allows you to write
$$ P(x) = Q(x)(ax + b) + R(x) $$
where the quotient is of degree 1 since $Q(x)$ has degree one less than $P(x)$, and
where the remainder $R(x)$ is of lesser degree than $Q(x)$. Dividing through by $Q(x)$ gives
$$ frac{P(x)}{Q(x)} = ax + b + frac{R(x)}{Q(x)}$$
Now if you let $x$ go to $infty$, the last term goes to 0, since the degree of $R(x)$ is less than the degree of $Q(x)$. IOW the LHS goes to $infty$ linearly i.e. just like $ax + b$ goes to $infty$. But the latter is a straight line with slope $a$.
That's the basic idea although you'll have to work a bit harder to "dot every i and cross every t."
answered Jan 15 at 21:55
NickDNickD
1,1211512
$endgroup$
add a comment |
$begingroup$
You are on the right track in the first case. Similar considerations can apply to the other two cases you examine.
If the degrees are equal, then both are growing at the same rate. In that case, factor out of each the highest power: e.g. if the numerator is $P(x) = ax^n + p(x)$ where $p(x)$ is of lower degree, then write it as $P(x) = ax^n (1 + frac{p(x)}{ax^n})$. Now, when $x$ goes to $infty$, the second term goes to $0$. Similarly, the denominator $Q(x) = bx^n + q(x)$ where $q(x)$ is of lower degree, can be written as $Q(x) = bx^n(1 + frac{q(x)}{bx^n})$ and the second term of this also goes to $0$ as $x$ goes to $infty$. You are left with $frac{P(x)}{Q(x)} = frac{ax^n}{bx^n} = frac{a}{b}$.
For the third case, remember that the polynomial division algorithm allows you to write
$$ P(x) = Q(x)(ax + b) + R(x) $$
where the quotient is of degree 1 since $Q(x)$ has degree one less than $P(x)$, and
where the remainder $R(x)$ is of lesser degree than $Q(x)$. Dividing through by $Q(x)$ gives
$$ frac{P(x)}{Q(x)} = ax + b + frac{R(x)}{Q(x)}$$
Now if you let $x$ go to $infty$, the last term goes to 0, since the degree of $R(x)$ is less than the degree of $Q(x)$. IOW the LHS goes to $infty$ linearly i.e. just like $ax + b$ goes to $infty$. But the latter is a straight line with slope $a$.
That's the basic idea although you'll have to work a bit harder to "dot every i and cross every t."
answered Jan 15 at 21:55
NickDNickD
1,1211512
$endgroup$
You are on the right track in the first case. Similar considerations can apply to the other two cases you examine.
If the degrees are equal, then both are growing at the same rate. In that case, factor out of each the highest power: e.g. if the numerator is $P(x) = ax^n + p(x)$ where $p(x)$ is of lower degree, then write it as $P(x) = ax^n (1 + frac{p(x)}{ax^n})$. Now, when $x$ goes to $infty$, the second term goes to $0$. Similarly, the denominator $Q(x) = bx^n + q(x)$ where $q(x)$ is of lower degree, can be written as $Q(x) = bx^n(1 + frac{q(x)}{bx^n})$ and the second term of this also goes to $0$ as $x$ goes to $infty$. You are left with $frac{P(x)}{Q(x)} = frac{ax^n}{bx^n} = frac{a}{b}$.
For the third case, remember that the polynomial division algorithm allows you to write
$$ P(x) = Q(x)(ax + b) + R(x) $$
where the quotient is of degree 1 since $Q(x)$ has degree one less than $P(x)$, and
where the remainder $R(x)$ is of lesser degree than $Q(x)$. Dividing through by $Q(x)$ gives
$$ frac{P(x)}{Q(x)} = ax + b + frac{R(x)}{Q(x)}$$
Now if you let $x$ go to $infty$, the last term goes to 0, since the degree of $R(x)$ is less than the degree of $Q(x)$. IOW the LHS goes to $infty$ linearly i.e. just like $ax + b$ goes to $infty$. But the latter is a straight line with slope $a$.
That's the basic idea although you'll have to work a bit harder to "dot every i and cross every t."
answered Jan 15 at 21:55
NickDNickD
1,1211512
edited Jan 15 at 22:40
answered Jan 15 at 21:55
NickDNickD
1,1211512
answered Jan 15 at 21:55
NickDNickD
1,1211512
answered Jan 15 at 21:55
NickDNickD
1,1211512
1,1211512
add a comment |
add a comment |
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