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












2












$begingroup$


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



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






calculus limits infinity







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$endgroup$








  • 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
    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
















2












$begingroup$


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



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






calculus limits infinity







share|cite|improve this question











$endgroup$








  • 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
    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














2












2








2


0



$begingroup$


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



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






calculus limits infinity







share|cite|improve this question











$endgroup$




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



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






calculus limits infinity




calculus limits infinity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 27 at 11:26









Asaf Karagila♦

306k33437767




306k33437767










asked Feb 27 at 9:48









user644361user644361

835




835








  • 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
    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




    $begingroup$
    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
    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 $color{red}0$. (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 $color{red}0$. (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










3 Answers
3






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











$endgroup$





















    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









    $endgroup$













    • $begingroup$
      Yeah I have understood that. That's why I have deleted my comment.
      $endgroup$
      – Dbchatto67
      Feb 27 at 10:18





















    0












    $begingroup$

    Let $-1/x=h$



    $$lim_{hto0^+}dfrac{dfracpi2-arctandfrac1h}{-h}=-lim_{...}dfrac{arctan h}h=-1$$



    See Are $mathrm{arccot}(x)$ and $arctan(1/x)$ the same function?






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      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











      $endgroup$


















        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











        $endgroup$
















          6












          6








          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











          $endgroup$



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 27 at 10:06

























          answered Feb 27 at 9:52









          5xum5xum

          91.3k394161




          91.3k394161























              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









              $endgroup$













              • $begingroup$
                Yeah I have understood that. That's why I have deleted my comment.
                $endgroup$
                – Dbchatto67
                Feb 27 at 10:18


















              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









              $endgroup$













              • $begingroup$
                Yeah I have understood that. That's why I have deleted my comment.
                $endgroup$
                – Dbchatto67
                Feb 27 at 10:18
















              2












              2








              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









              $endgroup$



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







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 27 at 9:51









              Kavi Rama MurthyKavi Rama Murthy

              65.1k42766




              65.1k42766












              • $begingroup$
                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






              $begingroup$
              Yeah I have understood that. That's why I have deleted my comment.
              $endgroup$
              – Dbchatto67
              Feb 27 at 10:18













              0












              $begingroup$

              Let $-1/x=h$



              $$lim_{hto0^+}dfrac{dfracpi2-arctandfrac1h}{-h}=-lim_{...}dfrac{arctan h}h=-1$$



              See Are $mathrm{arccot}(x)$ and $arctan(1/x)$ the same function?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let $-1/x=h$



                $$lim_{hto0^+}dfrac{dfracpi2-arctandfrac1h}{-h}=-lim_{...}dfrac{arctan h}h=-1$$



                See Are $mathrm{arccot}(x)$ and $arctan(1/x)$ the same function?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $-1/x=h$



                  $$lim_{hto0^+}dfrac{dfracpi2-arctandfrac1h}{-h}=-lim_{...}dfrac{arctan h}h=-1$$



                  See Are $mathrm{arccot}(x)$ and $arctan(1/x)$ the same function?






                  share|cite|improve this answer









                  $endgroup$



                  Let $-1/x=h$



                  $$lim_{hto0^+}dfrac{dfracpi2-arctandfrac1h}{-h}=-lim_{...}dfrac{arctan h}h=-1$$



                  See Are $mathrm{arccot}(x)$ and $arctan(1/x)$ the same function?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 27 at 15:00









                  lab bhattacharjeelab bhattacharjee

                  226k15157275




                  226k15157275






























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