Show that limit does not exist (two variables)












6












$begingroup$


I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.



I want to answer this question:




Show that the following limit does not exist



$$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2y^2+(x-y)^2}$$




So, my working:



$$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{xto 0}frac{0}{y^2}=0$$



and



$$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{yto 0}frac{0}{x^2}=0$$



I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $y=x$, not sure if that's correct - or my working.



So, let $y=x$, then



$$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2x^2+(x-y)^2}=lim_{xto 0}frac{x^4}{x^4+(x-x)^2}=1$$



Therefore limit doesn't exist.



Is this somewhat correct? What's the best way to answer a question like this?



Also, when showing that this limit does not exist, do I need to find different values of limits for both ${xto 0}$ AND ${yto 0}$? Or is one enough, e.g. if I just find two different values for two limits for ${xto 0}$ without using ${yto 0}$ in my calculation at all, is that fine?



Thanks!










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$endgroup$

















    6












    $begingroup$


    I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.



    I want to answer this question:




    Show that the following limit does not exist



    $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2y^2+(x-y)^2}$$




    So, my working:



    $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{xto 0}frac{0}{y^2}=0$$



    and



    $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{yto 0}frac{0}{x^2}=0$$



    I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $y=x$, not sure if that's correct - or my working.



    So, let $y=x$, then



    $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2x^2+(x-y)^2}=lim_{xto 0}frac{x^4}{x^4+(x-x)^2}=1$$



    Therefore limit doesn't exist.



    Is this somewhat correct? What's the best way to answer a question like this?



    Also, when showing that this limit does not exist, do I need to find different values of limits for both ${xto 0}$ AND ${yto 0}$? Or is one enough, e.g. if I just find two different values for two limits for ${xto 0}$ without using ${yto 0}$ in my calculation at all, is that fine?



    Thanks!










    share|cite|improve this question











    $endgroup$















      6












      6








      6


      0



      $begingroup$


      I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.



      I want to answer this question:




      Show that the following limit does not exist



      $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2y^2+(x-y)^2}$$




      So, my working:



      $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{xto 0}frac{0}{y^2}=0$$



      and



      $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{yto 0}frac{0}{x^2}=0$$



      I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $y=x$, not sure if that's correct - or my working.



      So, let $y=x$, then



      $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2x^2+(x-y)^2}=lim_{xto 0}frac{x^4}{x^4+(x-x)^2}=1$$



      Therefore limit doesn't exist.



      Is this somewhat correct? What's the best way to answer a question like this?



      Also, when showing that this limit does not exist, do I need to find different values of limits for both ${xto 0}$ AND ${yto 0}$? Or is one enough, e.g. if I just find two different values for two limits for ${xto 0}$ without using ${yto 0}$ in my calculation at all, is that fine?



      Thanks!










      share|cite|improve this question











      $endgroup$




      I just started looking into multiple variable calculus and limits involving them. I'm not amazing at limits either.



      I want to answer this question:




      Show that the following limit does not exist



      $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2y^2+(x-y)^2}$$




      So, my working:



      $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{xto 0}frac{0}{y^2}=0$$



      and



      $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}=lim_{yto 0}frac{0}{x^2}=0$$



      I wasn't sure what to do after this as they're both 0 but using the fact it can approach from any direction, I tried substituing $y=x$, not sure if that's correct - or my working.



      So, let $y=x$, then



      $$lim_{(x,y)to (0,0)}frac{x^2y^2}{x^2x^2+(x-y)^2}=lim_{xto 0}frac{x^4}{x^4+(x-x)^2}=1$$



      Therefore limit doesn't exist.



      Is this somewhat correct? What's the best way to answer a question like this?



      Also, when showing that this limit does not exist, do I need to find different values of limits for both ${xto 0}$ AND ${yto 0}$? Or is one enough, e.g. if I just find two different values for two limits for ${xto 0}$ without using ${yto 0}$ in my calculation at all, is that fine?



      Thanks!







      calculus limits multivariable-calculus proof-verification






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      share|cite|improve this question








      edited Feb 6 at 12:06









      José Carlos Santos

      161k22127232




      161k22127232










      asked Feb 6 at 11:44









      MandingoMandingo

      1434




      1434






















          5 Answers
          5






          active

          oldest

          votes


















          5












          $begingroup$

          What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same.



          So when you calculate $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.



          Similarly,
          $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.



          And the last limit you calculated is along line $y=x$.



          So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.






          share|cite|improve this answer









          $endgroup$









          • 5




            $begingroup$
            "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
            $endgroup$
            – leonbloy
            Feb 6 at 14:35






          • 2




            $begingroup$
            @leonbloy Do you have an example of that case?
            $endgroup$
            – Shufflepants
            Feb 6 at 15:48










          • $begingroup$
            @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
            $endgroup$
            – Swapnil Rustagi
            Feb 6 at 15:53








          • 2




            $begingroup$
            If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
            $endgroup$
            – gidds
            Feb 6 at 16:28










          • $begingroup$
            @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
            $endgroup$
            – leonbloy
            Feb 6 at 17:11





















          4












          $begingroup$

          Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How can I improve my reasoning?
            $endgroup$
            – Mandingo
            Feb 6 at 11:48






          • 1




            $begingroup$
            No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
            $endgroup$
            – PierreCarre
            Feb 6 at 11:54



















          1












          $begingroup$

          What you did is completely correct, but perhaps you can ease it a little bit as follows:



          For $;x=0;,;;yto0;$ the limit clearly is $;0;$, whereas for $;x=y;$ and $;xto0;$ we get



          $$frac{x^4}{x^4+(x-x)^2}=1xrightarrow[x=yto0]{}1$$



          Thus the limit depends on the path chosen $;implies;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You did right.



            A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
            $$
            lim_{xto0}frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
            lim_{xto0}frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
            lim_{xto0}frac{m^2}{m^2+(1-m)^2x^{-2}}=
            begin{cases}
            0 & mne 1 \[4px]
            1 & m=1
            end{cases}
            $$

            Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.






              share|cite|improve this answer











              $endgroup$













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                5 Answers
                5






                active

                oldest

                votes








                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                5












                $begingroup$

                What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same.



                So when you calculate $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.



                Similarly,
                $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.



                And the last limit you calculated is along line $y=x$.



                So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.






                share|cite|improve this answer









                $endgroup$









                • 5




                  $begingroup$
                  "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – leonbloy
                  Feb 6 at 14:35






                • 2




                  $begingroup$
                  @leonbloy Do you have an example of that case?
                  $endgroup$
                  – Shufflepants
                  Feb 6 at 15:48










                • $begingroup$
                  @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – Swapnil Rustagi
                  Feb 6 at 15:53








                • 2




                  $begingroup$
                  If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                  $endgroup$
                  – gidds
                  Feb 6 at 16:28










                • $begingroup$
                  @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                  $endgroup$
                  – leonbloy
                  Feb 6 at 17:11


















                5












                $begingroup$

                What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same.



                So when you calculate $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.



                Similarly,
                $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.



                And the last limit you calculated is along line $y=x$.



                So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.






                share|cite|improve this answer









                $endgroup$









                • 5




                  $begingroup$
                  "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – leonbloy
                  Feb 6 at 14:35






                • 2




                  $begingroup$
                  @leonbloy Do you have an example of that case?
                  $endgroup$
                  – Shufflepants
                  Feb 6 at 15:48










                • $begingroup$
                  @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – Swapnil Rustagi
                  Feb 6 at 15:53








                • 2




                  $begingroup$
                  If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                  $endgroup$
                  – gidds
                  Feb 6 at 16:28










                • $begingroup$
                  @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                  $endgroup$
                  – leonbloy
                  Feb 6 at 17:11
















                5












                5








                5





                $begingroup$

                What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same.



                So when you calculate $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.



                Similarly,
                $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.



                And the last limit you calculated is along line $y=x$.



                So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.






                share|cite|improve this answer









                $endgroup$



                What you have done is correct. The limit exists only if the value of the limit along every direction that leads to $(0,0)$ is same.



                So when you calculate $$lim_{xto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ you are calculating limit along the line $x=0$.



                Similarly,
                $$lim_{yto 0}frac{x^2y^2}{x^2y^2+(x-y)^2}$$ is limit along line $y=0$.



                And the last limit you calculated is along line $y=x$.



                So to answer your question, yes it would have been perfectly acceptable if you did not calculate limit along $y=0$. Just showing two examples where the limit comes out to be different along different directions is enough to show limit does not exist.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 6 at 11:56









                Swapnil RustagiSwapnil Rustagi

                741520




                741520








                • 5




                  $begingroup$
                  "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – leonbloy
                  Feb 6 at 14:35






                • 2




                  $begingroup$
                  @leonbloy Do you have an example of that case?
                  $endgroup$
                  – Shufflepants
                  Feb 6 at 15:48










                • $begingroup$
                  @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – Swapnil Rustagi
                  Feb 6 at 15:53








                • 2




                  $begingroup$
                  If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                  $endgroup$
                  – gidds
                  Feb 6 at 16:28










                • $begingroup$
                  @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                  $endgroup$
                  – leonbloy
                  Feb 6 at 17:11
















                • 5




                  $begingroup$
                  "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – leonbloy
                  Feb 6 at 14:35






                • 2




                  $begingroup$
                  @leonbloy Do you have an example of that case?
                  $endgroup$
                  – Shufflepants
                  Feb 6 at 15:48










                • $begingroup$
                  @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                  $endgroup$
                  – Swapnil Rustagi
                  Feb 6 at 15:53








                • 2




                  $begingroup$
                  If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                  $endgroup$
                  – gidds
                  Feb 6 at 16:28










                • $begingroup$
                  @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                  $endgroup$
                  – leonbloy
                  Feb 6 at 17:11










                5




                5




                $begingroup$
                "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                $endgroup$
                – leonbloy
                Feb 6 at 14:35




                $begingroup$
                "The limit exists only if the value of the limit along every direction ... is same" A word of caution: that's a necessary, but not a sufficient condition. It can happen that all directional limits give the same value, and the limit does not exist.
                $endgroup$
                – leonbloy
                Feb 6 at 14:35




                2




                2




                $begingroup$
                @leonbloy Do you have an example of that case?
                $endgroup$
                – Shufflepants
                Feb 6 at 15:48




                $begingroup$
                @leonbloy Do you have an example of that case?
                $endgroup$
                – Shufflepants
                Feb 6 at 15:48












                $begingroup$
                @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                $endgroup$
                – Swapnil Rustagi
                Feb 6 at 15:53






                $begingroup$
                @leonbloy I don't know of any way by which we can prove all directional limits exist and have same value, except by using epsilon-delta definition. Though I have very limited knowledge of multi-variable calculus. So it would be great if you could give an example where all directional limits give the same value, and the limit does not exist.
                $endgroup$
                – Swapnil Rustagi
                Feb 6 at 15:53






                2




                2




                $begingroup$
                If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                $endgroup$
                – gidds
                Feb 6 at 16:28




                $begingroup$
                If all directional limits give the same value, then the function might not be continuous at that point; it would need to be defined at that point, with a value equal to all the limits. But I think if all possible directional limits exist and are equal, then the function can be said to have that limit. (Unless anyone has a counterexample?)
                $endgroup$
                – gidds
                Feb 6 at 16:28












                $begingroup$
                @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                $endgroup$
                – leonbloy
                Feb 6 at 17:11






                $begingroup$
                @Shufflepants $f(x,y)=xy^2/(x^2+y^4)$ See here ocw.umb.edu/mathematics/math-240-calculus-iii-spring-2010/… (from page 19)
                $endgroup$
                – leonbloy
                Feb 6 at 17:11













                4












                $begingroup$

                Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  How can I improve my reasoning?
                  $endgroup$
                  – Mandingo
                  Feb 6 at 11:48






                • 1




                  $begingroup$
                  No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                  $endgroup$
                  – PierreCarre
                  Feb 6 at 11:54
















                4












                $begingroup$

                Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  How can I improve my reasoning?
                  $endgroup$
                  – Mandingo
                  Feb 6 at 11:48






                • 1




                  $begingroup$
                  No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                  $endgroup$
                  – PierreCarre
                  Feb 6 at 11:54














                4












                4








                4





                $begingroup$

                Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.






                share|cite|improve this answer









                $endgroup$



                Your reasoning is ok. If the limit existed, you would obtain the same value for every directional limit, which was not the case.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 6 at 11:47









                PierreCarrePierreCarre

                843111




                843111












                • $begingroup$
                  How can I improve my reasoning?
                  $endgroup$
                  – Mandingo
                  Feb 6 at 11:48






                • 1




                  $begingroup$
                  No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                  $endgroup$
                  – PierreCarre
                  Feb 6 at 11:54


















                • $begingroup$
                  How can I improve my reasoning?
                  $endgroup$
                  – Mandingo
                  Feb 6 at 11:48






                • 1




                  $begingroup$
                  No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                  $endgroup$
                  – PierreCarre
                  Feb 6 at 11:54
















                $begingroup$
                How can I improve my reasoning?
                $endgroup$
                – Mandingo
                Feb 6 at 11:48




                $begingroup$
                How can I improve my reasoning?
                $endgroup$
                – Mandingo
                Feb 6 at 11:48




                1




                1




                $begingroup$
                No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                $endgroup$
                – PierreCarre
                Feb 6 at 11:54




                $begingroup$
                No much room for improvement here... It is pretty standard. Maybe you could have just used directional limits instead of starting with the iterated limits, but in the end it is quite the same.
                $endgroup$
                – PierreCarre
                Feb 6 at 11:54











                1












                $begingroup$

                What you did is completely correct, but perhaps you can ease it a little bit as follows:



                For $;x=0;,;;yto0;$ the limit clearly is $;0;$, whereas for $;x=y;$ and $;xto0;$ we get



                $$frac{x^4}{x^4+(x-x)^2}=1xrightarrow[x=yto0]{}1$$



                Thus the limit depends on the path chosen $;implies;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  What you did is completely correct, but perhaps you can ease it a little bit as follows:



                  For $;x=0;,;;yto0;$ the limit clearly is $;0;$, whereas for $;x=y;$ and $;xto0;$ we get



                  $$frac{x^4}{x^4+(x-x)^2}=1xrightarrow[x=yto0]{}1$$



                  Thus the limit depends on the path chosen $;implies;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    What you did is completely correct, but perhaps you can ease it a little bit as follows:



                    For $;x=0;,;;yto0;$ the limit clearly is $;0;$, whereas for $;x=y;$ and $;xto0;$ we get



                    $$frac{x^4}{x^4+(x-x)^2}=1xrightarrow[x=yto0]{}1$$



                    Thus the limit depends on the path chosen $;implies;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.






                    share|cite|improve this answer









                    $endgroup$



                    What you did is completely correct, but perhaps you can ease it a little bit as follows:



                    For $;x=0;,;;yto0;$ the limit clearly is $;0;$, whereas for $;x=y;$ and $;xto0;$ we get



                    $$frac{x^4}{x^4+(x-x)^2}=1xrightarrow[x=yto0]{}1$$



                    Thus the limit depends on the path chosen $;implies;$ the limit doesn't exist...and this is the relevant argument: going on different paths yields different limits.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 6 at 11:55









                    DonAntonioDonAntonio

                    178k1494230




                    178k1494230























                        1












                        $begingroup$

                        You did right.



                        A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
                        $$
                        lim_{xto0}frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
                        lim_{xto0}frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
                        lim_{xto0}frac{m^2}{m^2+(1-m)^2x^{-2}}=
                        begin{cases}
                        0 & mne 1 \[4px]
                        1 & m=1
                        end{cases}
                        $$

                        Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You did right.



                          A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
                          $$
                          lim_{xto0}frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
                          lim_{xto0}frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
                          lim_{xto0}frac{m^2}{m^2+(1-m)^2x^{-2}}=
                          begin{cases}
                          0 & mne 1 \[4px]
                          1 & m=1
                          end{cases}
                          $$

                          Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You did right.



                            A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
                            $$
                            lim_{xto0}frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
                            lim_{xto0}frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
                            lim_{xto0}frac{m^2}{m^2+(1-m)^2x^{-2}}=
                            begin{cases}
                            0 & mne 1 \[4px]
                            1 & m=1
                            end{cases}
                            $$

                            Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.






                            share|cite|improve this answer









                            $endgroup$



                            You did right.



                            A possible improvement is to compute the limit along an arbitrary line: along $y=mx$ you have to compute
                            $$
                            lim_{xto0}frac{x^2(mx)^2}{x^2(mx)^2+(x-mx)^2}=
                            lim_{xto0}frac{m^2x^4}{m^2x^4+x^2(1-m)^2}=
                            lim_{xto0}frac{m^2}{m^2+(1-m)^2x^{-2}}=
                            begin{cases}
                            0 & mne 1 \[4px]
                            1 & m=1
                            end{cases}
                            $$

                            Doing this way may immediately show how to solve the business. Of course, if you find that all limits along lines are equal, you cannot conclude that the limit exists; instead, you should try along other curves, if you suspect the limit doesn't exist.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 6 at 14:10









                            egregegreg

                            182k1485203




                            182k1485203























                                1












                                $begingroup$

                                What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.






                                    share|cite|improve this answer











                                    $endgroup$



                                    What you did is correct, but note that after proving that the limit is $0$ when you take $y=0$, the fact that it is also $0$ when you take $x=0$ is irrelevant. What matters here is that going to $(0,0)$ through two different directions leads you to two distinct limits. Therefore, there is no (global) limit at $(0,0)$.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Feb 6 at 15:47

























                                    answered Feb 6 at 11:54









                                    José Carlos SantosJosé Carlos Santos

                                    161k22127232




                                    161k22127232






























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