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

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


How can I solve $lim_{xtoinfty}left(frac{2arctan(x)}{pi}right)^x$?Limit of $left(frac{-2}{pi}cdotarctan{x}right)^x$ when $xto -infty$, without using l'HôpitalDoes $f_{n}left(xright)=xarctanleft(nxright)$ converge uniformly?Why does Wolfram Alpha state that $-infty/0 = +infty$?Solving the limit $lim_{x to infty}frac{e^{frac{1}{x^2}}-1}{2arctan(x^2)-pi}$Find $lim_{xto infty} sinleft(frac 1xright) $As x approaches infinity, why does $ lim_{x to infty}arctan left(frac{x-2}{2}right) = frac{pi}{2} $Evaluate $lim_limits{x to +infty}left(xln (1+x)-xln x + arctanfrac{1}{2x}right)^{x^2arctan x}$Computing $lim_{xto -infty}xleft(frac{pi}{2}+arctan(x)right)$$lim_{nrightarrow infty} ( arctanfrac{1}{2} + arctan frac{1}{2.2^2} +…+ arctan frac {1}{2n^2})$













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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
    4 hours ago












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






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – Minus One-Twelfth
    4 hours ago
















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
    4 hours ago












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






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – Minus One-Twelfth
    4 hours ago














2












2








2


1



$begingroup$


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 3 hours ago









Asaf Karagila♦

305k33435766




305k33435766










asked 5 hours ago









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
    4 hours ago












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






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – Minus One-Twelfth
    4 hours ago














  • 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
    4 hours ago












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






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – Minus One-Twelfth
    4 hours ago








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
4 hours ago






$begingroup$
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
4 hours ago














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




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




1




1




$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
4 hours ago




$begingroup$
You're welcome!
$endgroup$
– Minus One-Twelfth
4 hours ago










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











$endgroup$





















    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
      4 hours ago













    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    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 4 hours ago

























        answered 5 hours ago









        5xum5xum

        91.2k394161




        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









            $endgroup$













            • $begingroup$
              Yeah I have understood that. That's why I have deleted my comment.
              $endgroup$
              – Dbchatto67
              4 hours ago


















            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
              4 hours ago
















            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 5 hours ago









            Kavi Rama MurthyKavi Rama Murthy

            63.2k42362




            63.2k42362












            • $begingroup$
              Yeah I have understood that. That's why I have deleted my comment.
              $endgroup$
              – Dbchatto67
              4 hours ago




















            • $begingroup$
              Yeah I have understood that. That's why I have deleted my comment.
              $endgroup$
              – Dbchatto67
              4 hours ago


















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






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




















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