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Something times negative infinity

what if take limit to negative infinity in the definition of e as a limitQuestion on limits and infinityHow does the chain rule for limits work?Why does Wolfram Alpha state that $-infty/0 = +infty$?negative and positive infinityWhich is correct: negative infinity or 'does not exist'?Finding a limit with negative infinity (Square root)What is infinity times negative number in Limit calculationNegative infinity equals positve infinity?Evaluating limits at positive and negative infinity

2

1

$begingroup$

Does $$lim_{x to - infty} (frac{pi}{2} + arctan{x} ) cdot x = - infty$$?

My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?

calculus limits infinity

share|cite|improve this question

asked 50 mins ago

user644361user644361

815

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The main flaw in your logic is that the "something" here happens to be $color{red}0$. (See points 2. and 3. of 5xum's answer, for example.)
  $endgroup$
  – Minus One-Twelfth
  26 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth Ah, I see it now. Thank you!
  $endgroup$
  – user644361
  25 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Minus One-Twelfth
  24 mins ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Does $$lim_{x to - infty} (frac{pi}{2} + arctan{x} ) cdot x = - infty$$?

My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?

calculus limits infinity

share|cite|improve this question

asked 50 mins ago

user644361user644361

815

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The main flaw in your logic is that the "something" here happens to be $color{red}0$. (See points 2. and 3. of 5xum's answer, for example.)
  $endgroup$
  – Minus One-Twelfth
  26 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth Ah, I see it now. Thank you!
  $endgroup$
  – user644361
  25 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Minus One-Twelfth
  24 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Does $$lim_{x to - infty} (frac{pi}{2} + arctan{x} ) cdot x = - infty$$?

My logic is that “something“ times "negative infinity" equals negative infinity. Am I right?

calculus limits infinity

share|cite|improve this question

asked 50 mins ago

user644361user644361

815

$endgroup$

share|cite|improve this question

asked 50 mins ago

user644361user644361

815

share|cite|improve this question

asked 50 mins ago

user644361user644361

815

asked 50 mins ago

user644361user644361

815

asked 50 mins ago

user644361user644361

815

2 Answers
2

active

oldest

votes

6

$begingroup$

The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:

1. 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?


2. Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_{xto-infty}alphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.


3. The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_{xto-infty} frac{1}{x}cdot x=-infty,$$ 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.

share|cite|improve this answer

edited 31 mins ago

answered 46 mins ago

5xum5xum

91.2k394161

$endgroup$

add a comment | 

2

$begingroup$

No. By L'Hopital's Rule $lim frac {frac {pi} 2+arctan, x} {1/x}=lim frac {1/(1+x^{2})} {-1/x^{2}}=-1$.

share|cite|improve this answer

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah I have understood that. That's why I have deleted my comment.
  $endgroup$
  – Dbchatto67
  20 mins ago
  
  
  

  
  
  
  

add a comment | 

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6

$begingroup$

The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:

1. 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?


2. Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_{xto-infty}alphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.


3. The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_{xto-infty} frac{1}{x}cdot x=-infty,$$ 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.

share|cite|improve this answer

edited 31 mins ago

answered 46 mins ago

5xum5xum

91.2k394161

$endgroup$

add a comment | 

6

$begingroup$

The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:

1. 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?


2. Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_{xto-infty}alphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.


3. The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_{xto-infty} frac{1}{x}cdot x=-infty,$$ 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.

share|cite|improve this answer

edited 31 mins ago

answered 46 mins ago

5xum5xum

91.2k394161

$endgroup$

add a comment | 

$begingroup$

The sentence "something times negative infinity equals negative infinity" is wrong in more than one way:

1. 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?


2. Even if you formalize the sentence, to something like "For all $alpha$, the limit $lim_{xto-infty}alphacdot x=-infty$, the statement remains false, since it is not true for any $alphaleq 0$.


3. The sentence, applied to another limit, shows how wrong it is to use it to calculate limits. By your logic, $$lim_{xto-infty} frac{1}{x}cdot x=-infty,$$ 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.

share|cite|improve this answer

edited 31 mins ago

answered 46 mins ago

5xum5xum

91.2k394161

$endgroup$

share|cite|improve this answer

edited 31 mins ago

answered 46 mins ago

5xum5xum

91.2k394161

answered 46 mins ago

5xum5xum

91.2k394161

answered 46 mins ago

5xum5xum

91.2k394161

2

$begingroup$

No. By L'Hopital's Rule $lim frac {frac {pi} 2+arctan, x} {1/x}=lim frac {1/(1+x^{2})} {-1/x^{2}}=-1$.

share|cite|improve this answer

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah I have understood that. That's why I have deleted my comment.
  $endgroup$
  – Dbchatto67
  20 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

No. By L'Hopital's Rule $lim frac {frac {pi} 2+arctan, x} {1/x}=lim frac {1/(1+x^{2})} {-1/x^{2}}=-1$.

share|cite|improve this answer

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah I have understood that. That's why I have deleted my comment.
  $endgroup$
  – Dbchatto67
  20 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

No. By L'Hopital's Rule $lim frac {frac {pi} 2+arctan, x} {1/x}=lim frac {1/(1+x^{2})} {-1/x^{2}}=-1$.

share|cite|improve this answer

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

$endgroup$

share|cite|improve this answer

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362

answered 46 mins ago

Kavi Rama MurthyKavi Rama Murthy

63.1k42362