Does $lim_x to - infty left(fracpi2 + arctanx right) cdot x = - infty$?
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Does $$lim_x to - infty left(fracpi2 + arctanx right) cdot x = - infty$$?
My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?
calculus limits infinity
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
$endgroup$
add a comment |
$begingroup$
Does $$lim_x to - infty left(fracpi2 + arctanx right) cdot x = - infty$$?
My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?
calculus limits infinity
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
$endgroup$
1
$begingroup$
The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
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– Minus One-Twelfth
Feb 27 at 10:11
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@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
1
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14
add a comment |
$begingroup$
Does $$lim_x to - infty left(fracpi2 + arctanx right) cdot x = - infty$$?
My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?
calculus limits infinity
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
$endgroup$
Does $$lim_x to - infty left(fracpi2 + arctanx right) cdot x = - infty$$?
My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?
calculus limits infinity
calculus limits infinity
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
edited Feb 27 at 11:26
Asaf Karagila♦
307k33439770
307k33439770
asked Feb 27 at 9:48
user644361user644361
855
asked Feb 27 at 9:48
user644361user644361
855
asked Feb 27 at 9:48
user644361user644361
855
855
1
$begingroup$
The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:11
$begingroup$
@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
1
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14
add a comment |
1
$begingroup$
The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:11
$begingroup$
@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
1
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14
1
1
$begingroup$
The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:11
$begingroup$
The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:11
$begingroup$
@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
$begingroup$
@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
1
1
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14
add a comment |
3 Answers
3
active
oldest
votes
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The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:
- It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
- Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_xto-inftyalphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.
- The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_xto-infty frac1xcdot x=-infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$
The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-infty$. Such a limit is often best approached using L'Hospital's rules.
answered Feb 27 at 9:52
5xum5xum
91.5k394161
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No. By L'Hopital's Rule $lim frac frac pi 2+arctan, x 1/x=lim frac 1/(1+x^2) -1/x^2=-1$.
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
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Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
add a comment |
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Let $-1/x=h$
$$lim_hto0^+dfracdfracpi2-arctandfrac1h-h=-lim_...dfracarctan hh=-1$$
See Are $mathrmarccot(x)$ and $arctan(1/x)$ the same function?
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:
- It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
- Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_xto-inftyalphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.
- The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_xto-infty frac1xcdot x=-infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$
The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-infty$. Such a limit is often best approached using L'Hospital's rules.
answered Feb 27 at 9:52
5xum5xum
91.5k394161
$endgroup$
add a comment |
$begingroup$
The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:
- It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
- Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_xto-inftyalphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.
- The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_xto-infty frac1xcdot x=-infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$
The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-infty$. Such a limit is often best approached using L'Hospital's rules.
answered Feb 27 at 9:52
5xum5xum
91.5k394161
$endgroup$
add a comment |
$begingroup$
The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:
- It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
- Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_xto-inftyalphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.
- The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_xto-infty frac1xcdot x=-infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$
The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-infty$. Such a limit is often best approached using L'Hospital's rules.
answered Feb 27 at 9:52
5xum5xum
91.5k394161
$endgroup$
The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:
- It is sloppy. Mathematics is rigorous for a reason. Mathematical statements must be well defined, and have a single meaning. In your case, what is "something"? Is an elephant times negative infinity equal to negative infinity?
- Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_xto-inftyalphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.
- The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_xto-infty frac1xcdot x=-infty,$$ since we have "something" times "negative infinity". Clearly, this is nonsence - the limit is obviously equal to $1$
The limit you must calculate is a limit of a product of two numbers, one tends to $0$, the other to $-infty$. Such a limit is often best approached using L'Hospital's rules.
answered Feb 27 at 9:52
5xum5xum
91.5k394161
edited Feb 27 at 10:06
answered Feb 27 at 9:52
5xum5xum
91.5k394161
answered Feb 27 at 9:52
5xum5xum
91.5k394161
answered Feb 27 at 9:52
5xum5xum
91.5k394161
91.5k394161
add a comment |
add a comment |
$begingroup$
No. By L'Hopital's Rule $lim frac frac pi 2+arctan, x 1/x=lim frac 1/(1+x^2) -1/x^2=-1$.
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
add a comment |
$begingroup$
No. By L'Hopital's Rule $lim frac frac pi 2+arctan, x 1/x=lim frac 1/(1+x^2) -1/x^2=-1$.
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
add a comment |
$begingroup$
No. By L'Hopital's Rule $lim frac frac pi 2+arctan, x 1/x=lim frac 1/(1+x^2) -1/x^2=-1$.
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
No. By L'Hopital's Rule $lim frac frac pi 2+arctan, x 1/x=lim frac 1/(1+x^2) -1/x^2=-1$.
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Feb 27 at 9:51
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
69.9k53170
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
add a comment |
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
$begingroup$
Yeah I have understood that. That's why I have deleted my comment.
$endgroup$
– Dbchatto67
Feb 27 at 10:18
add a comment |
$begingroup$
Let $-1/x=h$
$$lim_hto0^+dfracdfracpi2-arctandfrac1h-h=-lim_...dfracarctan hh=-1$$
See Are $mathrmarccot(x)$ and $arctan(1/x)$ the same function?
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
Let $-1/x=h$
$$lim_hto0^+dfracdfracpi2-arctandfrac1h-h=-lim_...dfracarctan hh=-1$$
See Are $mathrmarccot(x)$ and $arctan(1/x)$ the same function?
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
Let $-1/x=h$
$$lim_hto0^+dfracdfracpi2-arctandfrac1h-h=-lim_...dfracarctan hh=-1$$
See Are $mathrmarccot(x)$ and $arctan(1/x)$ the same function?
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
Let $-1/x=h$
$$lim_hto0^+dfracdfracpi2-arctandfrac1h-h=-lim_...dfracarctan hh=-1$$
See Are $mathrmarccot(x)$ and $arctan(1/x)$ the same function?
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
answered Feb 27 at 15:00
lab bhattacharjeelab bhattacharjee
227k15158278
227k15158278
add a comment |
add a comment |
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1
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The main flaw in your logic is that the "something" here happens to be $colorred0$. (See points 2. and 3. of 5xum's answer, for example.)
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:11
$begingroup$
@MinusOne-Twelfth Ah, I see it now. Thank you!
$endgroup$
– user644361
Feb 27 at 10:13
1
$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
Feb 27 at 10:14