How to find the other asymptote of $y=sqrt{x^2+x}$
$begingroup$
The task is to find the asymptotes of $y=sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $lim_{x to pm}y= infty$.
Next, to find $m_{1,2}$: $$m_1=lim_{x to +infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_1$$
and $$m_2=lim_{x to -infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=lim_{x to +infty}y-mx={sqrt{x^2+x} - x}=frac{(sqrt{x^2+x} - x)({sqrt{x^2+x} +x})}{{sqrt{x^2+x} +x}}=frac{x}{{sqrt{x^2+x} +x}}= frac{x}{x+{|x|sqrt{1+frac{1}{x}}}} to frac{x}{x+{|x|}}
approx frac{1}{1+frac{|x|}{x}}=frac{1}{2} $$
$$c_2=frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
real-analysis calculus limits asymptotics
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
$endgroup$
add a comment |
$begingroup$
The task is to find the asymptotes of $y=sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $lim_{x to pm}y= infty$.
Next, to find $m_{1,2}$: $$m_1=lim_{x to +infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_1$$
and $$m_2=lim_{x to -infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=lim_{x to +infty}y-mx={sqrt{x^2+x} - x}=frac{(sqrt{x^2+x} - x)({sqrt{x^2+x} +x})}{{sqrt{x^2+x} +x}}=frac{x}{{sqrt{x^2+x} +x}}= frac{x}{x+{|x|sqrt{1+frac{1}{x}}}} to frac{x}{x+{|x|}}
approx frac{1}{1+frac{|x|}{x}}=frac{1}{2} $$
$$c_2=frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
real-analysis calculus limits asymptotics
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
$endgroup$
1
$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54
add a comment |
$begingroup$
The task is to find the asymptotes of $y=sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $lim_{x to pm}y= infty$.
Next, to find $m_{1,2}$: $$m_1=lim_{x to +infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_1$$
and $$m_2=lim_{x to -infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=lim_{x to +infty}y-mx={sqrt{x^2+x} - x}=frac{(sqrt{x^2+x} - x)({sqrt{x^2+x} +x})}{{sqrt{x^2+x} +x}}=frac{x}{{sqrt{x^2+x} +x}}= frac{x}{x+{|x|sqrt{1+frac{1}{x}}}} to frac{x}{x+{|x|}}
approx frac{1}{1+frac{|x|}{x}}=frac{1}{2} $$
$$c_2=frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
real-analysis calculus limits asymptotics
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
$endgroup$
The task is to find the asymptotes of $y=sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $lim_{x to pm}y= infty$.
Next, to find $m_{1,2}$: $$m_1=lim_{x to +infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_1$$
and $$m_2=lim_{x to -infty}frac{y}{x}=frac{sqrt{x^2+x}}{x}=sqrt{1+ frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=lim_{x to +infty}y-mx={sqrt{x^2+x} - x}=frac{(sqrt{x^2+x} - x)({sqrt{x^2+x} +x})}{{sqrt{x^2+x} +x}}=frac{x}{{sqrt{x^2+x} +x}}= frac{x}{x+{|x|sqrt{1+frac{1}{x}}}} to frac{x}{x+{|x|}}
approx frac{1}{1+frac{|x|}{x}}=frac{1}{2} $$
$$c_2=frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
real-analysis calculus limits asymptotics
real-analysis calculus limits asymptotics
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
asked Dec 19 '18 at 16:14
E.NoleE.Nole
139114
139114
1
$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54
add a comment |
1
$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54
1
1
$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54
$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $xto pminfty$ if $f(x)=mx+q+o(1)$. Now, as $|x|to +infty$,
$$sqrt{x^2+x}=|x|sqrt{1+1/x}=|x|left(1+frac{1}{2x}+o(1/x)right)\
=|x|+frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$sqrt{x^2+x}=x+frac{1}{2}+o(1)$$
and for $x<0$
$$sqrt{x^2+x}=-x-frac{1}{2}+o(1).$$
What may we conclude?
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
$endgroup$
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
add a comment |
$begingroup$
You have the wrong value for $m_2.$
If $x leq -1,$ then $frac{sqrt{x^2+x}}{x} neq sqrt{1+ frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $xto pminfty$ if $f(x)=mx+q+o(1)$. Now, as $|x|to +infty$,
$$sqrt{x^2+x}=|x|sqrt{1+1/x}=|x|left(1+frac{1}{2x}+o(1/x)right)\
=|x|+frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$sqrt{x^2+x}=x+frac{1}{2}+o(1)$$
and for $x<0$
$$sqrt{x^2+x}=-x-frac{1}{2}+o(1).$$
What may we conclude?
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
$endgroup$
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
add a comment |
$begingroup$
You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $xto pminfty$ if $f(x)=mx+q+o(1)$. Now, as $|x|to +infty$,
$$sqrt{x^2+x}=|x|sqrt{1+1/x}=|x|left(1+frac{1}{2x}+o(1/x)right)\
=|x|+frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$sqrt{x^2+x}=x+frac{1}{2}+o(1)$$
and for $x<0$
$$sqrt{x^2+x}=-x-frac{1}{2}+o(1).$$
What may we conclude?
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
$endgroup$
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
add a comment |
$begingroup$
You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $xto pminfty$ if $f(x)=mx+q+o(1)$. Now, as $|x|to +infty$,
$$sqrt{x^2+x}=|x|sqrt{1+1/x}=|x|left(1+frac{1}{2x}+o(1/x)right)\
=|x|+frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$sqrt{x^2+x}=x+frac{1}{2}+o(1)$$
and for $x<0$
$$sqrt{x^2+x}=-x-frac{1}{2}+o(1).$$
What may we conclude?
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
$endgroup$
You may try also in this way. By definition $y=mx+q$ is an asymptote of $f$ as $xto pminfty$ if $f(x)=mx+q+o(1)$. Now, as $|x|to +infty$,
$$sqrt{x^2+x}=|x|sqrt{1+1/x}=|x|left(1+frac{1}{2x}+o(1/x)right)\
=|x|+frac{|x|}{2x}+o(1).$$
Hence if $x>0$ then
$$sqrt{x^2+x}=x+frac{1}{2}+o(1)$$
and for $x<0$
$$sqrt{x^2+x}=-x-frac{1}{2}+o(1).$$
What may we conclude?
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
edited Dec 19 '18 at 16:35
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
answered Dec 19 '18 at 16:25
Robert ZRobert Z
95.8k1065136
95.8k1065136
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
add a comment |
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
$begingroup$
Well, since you have used asymptotic expansions, I think this means that if when $xto pminfty$ we have that $f(x)=mx+c+o(1)$ then this implies that $y=mx+c$ is an asymptote? Is this the right conclusion? Pardon me, but my analysis textbook doesn't talk about asymptotics so maybe I'm not applying the definition properly. Therefore, if you may as well sir, please advise what book or resource I can use to learn how to use asymptotics in calculating limits. I really like your approach.
$endgroup$
– E.Nole
Dec 19 '18 at 16:49
1
1
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
$begingroup$
Yes, that's right. Maybe my approach is unusual at this level (and I do not have a a reference to suggest). However you can read the definition here: en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes
$endgroup$
– Robert Z
Dec 19 '18 at 18:31
add a comment |
$begingroup$
You have the wrong value for $m_2.$
If $x leq -1,$ then $frac{sqrt{x^2+x}}{x} neq sqrt{1+ frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
$endgroup$
add a comment |
$begingroup$
You have the wrong value for $m_2.$
If $x leq -1,$ then $frac{sqrt{x^2+x}}{x} neq sqrt{1+ frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
$endgroup$
add a comment |
$begingroup$
You have the wrong value for $m_2.$
If $x leq -1,$ then $frac{sqrt{x^2+x}}{x} neq sqrt{1+ frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
$endgroup$
You have the wrong value for $m_2.$
If $x leq -1,$ then $frac{sqrt{x^2+x}}{x} neq sqrt{1+ frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
answered Dec 19 '18 at 16:22
David KDavid K
53.8k342116
53.8k342116
add a comment |
add a comment |
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$begingroup$
$sqrt{x^2}=-x$ if $x<0$.
$endgroup$
– egreg
Dec 19 '18 at 16:54