Limit to 0 ambiguity The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists

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Limit to 0 ambiguity



The 2019 Stack Overflow Developer Survey Results Are InLimit Calculation - Sequences at infinityNeed help with a limitStrategy for tackling $lim_ntoinftyfracn(ln n)^-p.$Evaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Finding a basic limitCan the limit of a polynomial involving infinity be finite?Limit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists










1












$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago















1












$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago













1












1








1





$begingroup$


I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.






limits







share|cite|improve this question











$endgroup$




I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 mins ago









user8718165

1167




1167










asked 1 hour ago









J.MohJ.Moh

475




475







  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago












  • 1




    $begingroup$
    Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
    $endgroup$
    – Dave
    1 hour ago






  • 3




    $begingroup$
    How are you defining a limit?
    $endgroup$
    – John Doe
    1 hour ago










  • $begingroup$
    Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
    $endgroup$
    – J.Moh
    1 hour ago







1




1




$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
1 hour ago




$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
1 hour ago




3




3




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago




$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
1 hour ago












$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago




$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



    enter image description here



    As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



    $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



    For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



    Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – J.Moh
      37 mins ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.






        share|cite|improve this answer









        $endgroup$



        The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        MPWMPW

        31.2k12157




        31.2k12157





















            2












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              37 mins ago















            2












            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              37 mins ago













            2












            2








            2





            $begingroup$

            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.






            share|cite|improve this answer











            $endgroup$



            It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:



            enter image description here



            As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted as $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted as $xto 0^+$), the function grows without bound positively. Analytically, this fact is written as follows:



            $$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$



            For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).



            Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 49 mins ago









            J. W. Tanner

            4,6441320




            4,6441320










            answered 1 hour ago









            Michael RybkinMichael Rybkin

            4,259422




            4,259422











            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              37 mins ago
















            • $begingroup$
              Thank you so much!!!
              $endgroup$
              – J.Moh
              37 mins ago















            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            37 mins ago




            $begingroup$
            Thank you so much!!!
            $endgroup$
            – J.Moh
            37 mins ago

















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