Prove that $lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=0$.












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$begingroup$



Prove that $displaystyle lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=0$.




This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...



$displaystyle|f(x,y)-L|=left |frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}} right | leq frac{|x^{2}+xy+y^{2}|}{big|sqrt{x^{2}+y^{2}}big|}cdot frac{sqrt{x^{2}+y^{2}}}{sqrt{x^{2}+y^{2}}} = frac{big(sqrt{x^{2}+y^{2}}big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$



I wanted the square root on top to help determine what I should set my $delta$ as. Any ideas on how to finish this or do it in a better way?






limits multivariable-calculus epsilon-delta







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$endgroup$

















    2












    $begingroup$



    Prove that $displaystyle lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=0$.




    This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...



    $displaystyle|f(x,y)-L|=left |frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}} right | leq frac{|x^{2}+xy+y^{2}|}{big|sqrt{x^{2}+y^{2}}big|}cdot frac{sqrt{x^{2}+y^{2}}}{sqrt{x^{2}+y^{2}}} = frac{big(sqrt{x^{2}+y^{2}}big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$



    I wanted the square root on top to help determine what I should set my $delta$ as. Any ideas on how to finish this or do it in a better way?






    limits multivariable-calculus epsilon-delta







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Prove that $displaystyle lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=0$.




      This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...



      $displaystyle|f(x,y)-L|=left |frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}} right | leq frac{|x^{2}+xy+y^{2}|}{big|sqrt{x^{2}+y^{2}}big|}cdot frac{sqrt{x^{2}+y^{2}}}{sqrt{x^{2}+y^{2}}} = frac{big(sqrt{x^{2}+y^{2}}big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$



      I wanted the square root on top to help determine what I should set my $delta$ as. Any ideas on how to finish this or do it in a better way?






      limits multivariable-calculus epsilon-delta







      share|cite|improve this question











      $endgroup$





      Prove that $displaystyle lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=0$.




      This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...



      $displaystyle|f(x,y)-L|=left |frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}} right | leq frac{|x^{2}+xy+y^{2}|}{big|sqrt{x^{2}+y^{2}}big|}cdot frac{sqrt{x^{2}+y^{2}}}{sqrt{x^{2}+y^{2}}} = frac{big(sqrt{x^{2}+y^{2}}big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$



      I wanted the square root on top to help determine what I should set my $delta$ as. Any ideas on how to finish this or do it in a better way?






      limits multivariable-calculus epsilon-delta




      limits multivariable-calculus epsilon-delta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 3 at 6:04









      El borito

      664216




      664216










      asked Feb 3 at 3:32









      numericalorangenumericalorange

      1,949314




      1,949314






















          4 Answers
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          2












          $begingroup$

          Hint: To finish ...



          $$0 leqslantfrac{(sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}leqslant sqrt{x^{2}+y^{2}}left(frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+frac{|xy|}{|x^{2}+y^{2}|}right) leqslant sqrt{x^{2}+y^{2}}(ldots)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Another hint: $(|x| - |y|)^2 geqslant 0$
            $endgroup$
            – RRL
            Feb 3 at 3:48










          • $begingroup$
            Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
            $endgroup$
            – numericalorange
            Feb 3 at 4:04












          • $begingroup$
            I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
            $endgroup$
            – numericalorange
            Feb 3 at 4:05






          • 1




            $begingroup$
            So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
            $endgroup$
            – RRL
            Feb 3 at 4:07






          • 1




            $begingroup$
            ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
            $endgroup$
            – RRL
            Feb 3 at 4:09



















          2












          $begingroup$

          To begin with, notice that
          begin{align*}
          begin{cases}
          |x| = sqrt{x^{2}} leq sqrt{x^{2}+y^{2}}\\
          |y| = sqrt{y^{2}} leq sqrt{x^{2}+y^{2}}
          end{cases}Longrightarrow
          begin{cases}
          displaystylefrac{|x|}{sqrt{x^{2}+y^{2}}} leq 1\\
          displaystylefrac{|y|}{sqrt{x^{2}+y^{2}}} leq 1
          end{cases}
          end{align*}
          Furthermore, the given expression can be split as the following sum



          begin{align*}
          E(x,y) = frac{x^{2}+2xy+y^{2}}{sqrt{x^{2}+y^{2}}} = frac{x^{2}}{sqrt{x^{2}+y^{2}}} + frac{2xy}{sqrt{x^{2}+y^{2}}} + frac{y^{2}}{sqrt{x^{2} + y^{2}}}
          end{align*}



          As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You can use polar coordinates:
            $$x=rcos t; y=r sin t;\
            lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=lim_{rto 0}frac{r^{2}(1+frac12sin 2t)}{|r|}=0.$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):




              • $sqrt{ab}leq sqrt{frac{a^2+b^2}{2}}$


              So, you get
              begin{eqnarray*} left| frac{x^2+xy+y^2}{sqrt{x^2+y^2}}right|
              & = & left| frac{x^2+y^2}{sqrt{x^2+y^2}} + frac{xy}{sqrt{x^2+y^2}}right| \
              & leq & sqrt{x^2+y^2} + frac{|xy|}{sqrt{x^2+y^2}}\
              & stackrel{GM-QM}{leq} & sqrt{x^2+y^2} + frac{left(sqrt{frac{x^2+y^2}{2}} right)^2}{sqrt{x^2+y^2}} \
              & = & frac{3}{2}sqrt{x^2+y^2} \
              & stackrel{(x,y)to (0,0)}{longrightarrow} & 0
              end{eqnarray*}






              share|cite|improve this answer









              $endgroup$














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                4 Answers
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                4 Answers
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                votes






                active

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                2












                $begingroup$

                Hint: To finish ...



                $$0 leqslantfrac{(sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}leqslant sqrt{x^{2}+y^{2}}left(frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+frac{|xy|}{|x^{2}+y^{2}|}right) leqslant sqrt{x^{2}+y^{2}}(ldots)$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Another hint: $(|x| - |y|)^2 geqslant 0$
                  $endgroup$
                  – RRL
                  Feb 3 at 3:48










                • $begingroup$
                  Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:04












                • $begingroup$
                  I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:05






                • 1




                  $begingroup$
                  So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:07






                • 1




                  $begingroup$
                  ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:09
















                2












                $begingroup$

                Hint: To finish ...



                $$0 leqslantfrac{(sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}leqslant sqrt{x^{2}+y^{2}}left(frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+frac{|xy|}{|x^{2}+y^{2}|}right) leqslant sqrt{x^{2}+y^{2}}(ldots)$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Another hint: $(|x| - |y|)^2 geqslant 0$
                  $endgroup$
                  – RRL
                  Feb 3 at 3:48










                • $begingroup$
                  Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:04












                • $begingroup$
                  I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:05






                • 1




                  $begingroup$
                  So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:07






                • 1




                  $begingroup$
                  ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:09














                2












                2








                2





                $begingroup$

                Hint: To finish ...



                $$0 leqslantfrac{(sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}leqslant sqrt{x^{2}+y^{2}}left(frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+frac{|xy|}{|x^{2}+y^{2}|}right) leqslant sqrt{x^{2}+y^{2}}(ldots)$$






                share|cite|improve this answer









                $endgroup$



                Hint: To finish ...



                $$0 leqslantfrac{(sqrt{x^{2}+y^{2}})|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}leqslant sqrt{x^{2}+y^{2}}left(frac{|x^{2}+y^{2}|}{|x^{2}+y^{2}|}+frac{|xy|}{|x^{2}+y^{2}|}right) leqslant sqrt{x^{2}+y^{2}}(ldots)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 3 at 3:41









                RRLRRL

                53.7k52574




                53.7k52574












                • $begingroup$
                  Another hint: $(|x| - |y|)^2 geqslant 0$
                  $endgroup$
                  – RRL
                  Feb 3 at 3:48










                • $begingroup$
                  Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:04












                • $begingroup$
                  I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:05






                • 1




                  $begingroup$
                  So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:07






                • 1




                  $begingroup$
                  ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:09


















                • $begingroup$
                  Another hint: $(|x| - |y|)^2 geqslant 0$
                  $endgroup$
                  – RRL
                  Feb 3 at 3:48










                • $begingroup$
                  Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:04












                • $begingroup$
                  I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                  $endgroup$
                  – numericalorange
                  Feb 3 at 4:05






                • 1




                  $begingroup$
                  So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:07






                • 1




                  $begingroup$
                  ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                  $endgroup$
                  – RRL
                  Feb 3 at 4:09
















                $begingroup$
                Another hint: $(|x| - |y|)^2 geqslant 0$
                $endgroup$
                – RRL
                Feb 3 at 3:48




                $begingroup$
                Another hint: $(|x| - |y|)^2 geqslant 0$
                $endgroup$
                – RRL
                Feb 3 at 3:48












                $begingroup$
                Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                $endgroup$
                – numericalorange
                Feb 3 at 4:04






                $begingroup$
                Thank you, I've been trying to do your hint. I got so far $(sqrt{x^{2}+y^{2}})(1+frac{|x|^{2}-2|xy|+|y|^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(|x|-|y|)^{2}}{|x^{2}+y^{2}|}) leq (sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})$ ...is this correct? I am not sure how to finish this!
                $endgroup$
                – numericalorange
                Feb 3 at 4:04














                $begingroup$
                I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                $endgroup$
                – numericalorange
                Feb 3 at 4:05




                $begingroup$
                I then did: $(sqrt{x^{2}+y^{2}})(1+frac{(x^{2}-y^{2})(x^{2}+y^{2})}{|x^{2}+y^{2}|})leq (sqrt{x^{2}+y^{2}})(1+x^{2}-y^{2})leq (sqrt{x^{2}+y^{2}})(1+x^{2})$...
                $endgroup$
                – numericalorange
                Feb 3 at 4:05




                1




                1




                $begingroup$
                So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                $endgroup$
                – RRL
                Feb 3 at 4:07




                $begingroup$
                So $(|x| - |y|)^2 geqslant 0 implies x^2 -2|xy| + y^2 geqslant 0 implies x^2 + y^2 geqslant 2|xy| geqslant |xy|$.
                $endgroup$
                – RRL
                Feb 3 at 4:07




                1




                1




                $begingroup$
                ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                $endgroup$
                – RRL
                Feb 3 at 4:09




                $begingroup$
                ... and, of course $frac{|xy|}{x^2 + y^2} leqslant 1$ Thus the bound on the right is $2sqrt{x^2 + y^2}$. Now you proceed with $epsilon- delta$ or just use the squeeze theorem.
                $endgroup$
                – RRL
                Feb 3 at 4:09











                2












                $begingroup$

                To begin with, notice that
                begin{align*}
                begin{cases}
                |x| = sqrt{x^{2}} leq sqrt{x^{2}+y^{2}}\\
                |y| = sqrt{y^{2}} leq sqrt{x^{2}+y^{2}}
                end{cases}Longrightarrow
                begin{cases}
                displaystylefrac{|x|}{sqrt{x^{2}+y^{2}}} leq 1\\
                displaystylefrac{|y|}{sqrt{x^{2}+y^{2}}} leq 1
                end{cases}
                end{align*}
                Furthermore, the given expression can be split as the following sum



                begin{align*}
                E(x,y) = frac{x^{2}+2xy+y^{2}}{sqrt{x^{2}+y^{2}}} = frac{x^{2}}{sqrt{x^{2}+y^{2}}} + frac{2xy}{sqrt{x^{2}+y^{2}}} + frac{y^{2}}{sqrt{x^{2} + y^{2}}}
                end{align*}



                As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  To begin with, notice that
                  begin{align*}
                  begin{cases}
                  |x| = sqrt{x^{2}} leq sqrt{x^{2}+y^{2}}\\
                  |y| = sqrt{y^{2}} leq sqrt{x^{2}+y^{2}}
                  end{cases}Longrightarrow
                  begin{cases}
                  displaystylefrac{|x|}{sqrt{x^{2}+y^{2}}} leq 1\\
                  displaystylefrac{|y|}{sqrt{x^{2}+y^{2}}} leq 1
                  end{cases}
                  end{align*}
                  Furthermore, the given expression can be split as the following sum



                  begin{align*}
                  E(x,y) = frac{x^{2}+2xy+y^{2}}{sqrt{x^{2}+y^{2}}} = frac{x^{2}}{sqrt{x^{2}+y^{2}}} + frac{2xy}{sqrt{x^{2}+y^{2}}} + frac{y^{2}}{sqrt{x^{2} + y^{2}}}
                  end{align*}



                  As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    To begin with, notice that
                    begin{align*}
                    begin{cases}
                    |x| = sqrt{x^{2}} leq sqrt{x^{2}+y^{2}}\\
                    |y| = sqrt{y^{2}} leq sqrt{x^{2}+y^{2}}
                    end{cases}Longrightarrow
                    begin{cases}
                    displaystylefrac{|x|}{sqrt{x^{2}+y^{2}}} leq 1\\
                    displaystylefrac{|y|}{sqrt{x^{2}+y^{2}}} leq 1
                    end{cases}
                    end{align*}
                    Furthermore, the given expression can be split as the following sum



                    begin{align*}
                    E(x,y) = frac{x^{2}+2xy+y^{2}}{sqrt{x^{2}+y^{2}}} = frac{x^{2}}{sqrt{x^{2}+y^{2}}} + frac{2xy}{sqrt{x^{2}+y^{2}}} + frac{y^{2}}{sqrt{x^{2} + y^{2}}}
                    end{align*}



                    As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.






                    share|cite|improve this answer









                    $endgroup$



                    To begin with, notice that
                    begin{align*}
                    begin{cases}
                    |x| = sqrt{x^{2}} leq sqrt{x^{2}+y^{2}}\\
                    |y| = sqrt{y^{2}} leq sqrt{x^{2}+y^{2}}
                    end{cases}Longrightarrow
                    begin{cases}
                    displaystylefrac{|x|}{sqrt{x^{2}+y^{2}}} leq 1\\
                    displaystylefrac{|y|}{sqrt{x^{2}+y^{2}}} leq 1
                    end{cases}
                    end{align*}
                    Furthermore, the given expression can be split as the following sum



                    begin{align*}
                    E(x,y) = frac{x^{2}+2xy+y^{2}}{sqrt{x^{2}+y^{2}}} = frac{x^{2}}{sqrt{x^{2}+y^{2}}} + frac{2xy}{sqrt{x^{2}+y^{2}}} + frac{y^{2}}{sqrt{x^{2} + y^{2}}}
                    end{align*}



                    As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 3 at 3:50









                    APC89APC89

                    2,371720




                    2,371720























                        1












                        $begingroup$

                        You can use polar coordinates:
                        $$x=rcos t; y=r sin t;\
                        lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=lim_{rto 0}frac{r^{2}(1+frac12sin 2t)}{|r|}=0.$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You can use polar coordinates:
                          $$x=rcos t; y=r sin t;\
                          lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=lim_{rto 0}frac{r^{2}(1+frac12sin 2t)}{|r|}=0.$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You can use polar coordinates:
                            $$x=rcos t; y=r sin t;\
                            lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=lim_{rto 0}frac{r^{2}(1+frac12sin 2t)}{|r|}=0.$$






                            share|cite|improve this answer









                            $endgroup$



                            You can use polar coordinates:
                            $$x=rcos t; y=r sin t;\
                            lim_{(x,y)to (0,0)}frac{x^{2}+xy+y^{2}}{sqrt{x^{2}+y^{2}}}=lim_{rto 0}frac{r^{2}(1+frac12sin 2t)}{|r|}=0.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 3 at 6:38









                            farruhotafarruhota

                            22.2k2942




                            22.2k2942























                                1












                                $begingroup$

                                You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):




                                • $sqrt{ab}leq sqrt{frac{a^2+b^2}{2}}$


                                So, you get
                                begin{eqnarray*} left| frac{x^2+xy+y^2}{sqrt{x^2+y^2}}right|
                                & = & left| frac{x^2+y^2}{sqrt{x^2+y^2}} + frac{xy}{sqrt{x^2+y^2}}right| \
                                & leq & sqrt{x^2+y^2} + frac{|xy|}{sqrt{x^2+y^2}}\
                                & stackrel{GM-QM}{leq} & sqrt{x^2+y^2} + frac{left(sqrt{frac{x^2+y^2}{2}} right)^2}{sqrt{x^2+y^2}} \
                                & = & frac{3}{2}sqrt{x^2+y^2} \
                                & stackrel{(x,y)to (0,0)}{longrightarrow} & 0
                                end{eqnarray*}






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):




                                  • $sqrt{ab}leq sqrt{frac{a^2+b^2}{2}}$


                                  So, you get
                                  begin{eqnarray*} left| frac{x^2+xy+y^2}{sqrt{x^2+y^2}}right|
                                  & = & left| frac{x^2+y^2}{sqrt{x^2+y^2}} + frac{xy}{sqrt{x^2+y^2}}right| \
                                  & leq & sqrt{x^2+y^2} + frac{|xy|}{sqrt{x^2+y^2}}\
                                  & stackrel{GM-QM}{leq} & sqrt{x^2+y^2} + frac{left(sqrt{frac{x^2+y^2}{2}} right)^2}{sqrt{x^2+y^2}} \
                                  & = & frac{3}{2}sqrt{x^2+y^2} \
                                  & stackrel{(x,y)to (0,0)}{longrightarrow} & 0
                                  end{eqnarray*}






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):




                                    • $sqrt{ab}leq sqrt{frac{a^2+b^2}{2}}$


                                    So, you get
                                    begin{eqnarray*} left| frac{x^2+xy+y^2}{sqrt{x^2+y^2}}right|
                                    & = & left| frac{x^2+y^2}{sqrt{x^2+y^2}} + frac{xy}{sqrt{x^2+y^2}}right| \
                                    & leq & sqrt{x^2+y^2} + frac{|xy|}{sqrt{x^2+y^2}}\
                                    & stackrel{GM-QM}{leq} & sqrt{x^2+y^2} + frac{left(sqrt{frac{x^2+y^2}{2}} right)^2}{sqrt{x^2+y^2}} \
                                    & = & frac{3}{2}sqrt{x^2+y^2} \
                                    & stackrel{(x,y)to (0,0)}{longrightarrow} & 0
                                    end{eqnarray*}






                                    share|cite|improve this answer









                                    $endgroup$



                                    You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):




                                    • $sqrt{ab}leq sqrt{frac{a^2+b^2}{2}}$


                                    So, you get
                                    begin{eqnarray*} left| frac{x^2+xy+y^2}{sqrt{x^2+y^2}}right|
                                    & = & left| frac{x^2+y^2}{sqrt{x^2+y^2}} + frac{xy}{sqrt{x^2+y^2}}right| \
                                    & leq & sqrt{x^2+y^2} + frac{|xy|}{sqrt{x^2+y^2}}\
                                    & stackrel{GM-QM}{leq} & sqrt{x^2+y^2} + frac{left(sqrt{frac{x^2+y^2}{2}} right)^2}{sqrt{x^2+y^2}} \
                                    & = & frac{3}{2}sqrt{x^2+y^2} \
                                    & stackrel{(x,y)to (0,0)}{longrightarrow} & 0
                                    end{eqnarray*}







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Feb 3 at 7:07









                                    trancelocationtrancelocation

                                    14.1k1829




                                    14.1k1829






























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