Does $lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0$?












3















Does $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 ?$$




I'm trying to show that $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 $$



but I am getting stuck. I was thinking that as a starting point I could show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0 $$



and then conclude that since $$lim_{x to 0} sqrt{x} = 0$$ and $frac{a^2b^2}{a^2 + b^2} to 0$ as $(a, b) to (0, 0)$ we arrive at $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0.$$




Firstly is my approach above a correct one. Secondly how can show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.






limits multivariable-calculus







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  • $limsqrt{f}=sqrt{lim f}$ if the limit exists
    – Nosrati
    Nov 24 at 14:49
















3















Does $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 ?$$




I'm trying to show that $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 $$



but I am getting stuck. I was thinking that as a starting point I could show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0 $$



and then conclude that since $$lim_{x to 0} sqrt{x} = 0$$ and $frac{a^2b^2}{a^2 + b^2} to 0$ as $(a, b) to (0, 0)$ we arrive at $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0.$$




Firstly is my approach above a correct one. Secondly how can show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.






limits multivariable-calculus







share|cite|improve this question






















  • $limsqrt{f}=sqrt{lim f}$ if the limit exists
    – Nosrati
    Nov 24 at 14:49














3












3








3








Does $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 ?$$




I'm trying to show that $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 $$



but I am getting stuck. I was thinking that as a starting point I could show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0 $$



and then conclude that since $$lim_{x to 0} sqrt{x} = 0$$ and $frac{a^2b^2}{a^2 + b^2} to 0$ as $(a, b) to (0, 0)$ we arrive at $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0.$$




Firstly is my approach above a correct one. Secondly how can show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.






limits multivariable-calculus







share|cite|improve this question














Does $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 ?$$




I'm trying to show that $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0 $$



but I am getting stuck. I was thinking that as a starting point I could show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0 $$



and then conclude that since $$lim_{x to 0} sqrt{x} = 0$$ and $frac{a^2b^2}{a^2 + b^2} to 0$ as $(a, b) to (0, 0)$ we arrive at $$lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} = 0.$$




Firstly is my approach above a correct one. Secondly how can show that $$lim_{(a, b) to (0, 0)} frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.






limits multivariable-calculus




limits multivariable-calculus






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asked Nov 24 at 14:47









Perturbative

4,07011449




4,07011449












  • $limsqrt{f}=sqrt{lim f}$ if the limit exists
    – Nosrati
    Nov 24 at 14:49


















  • $limsqrt{f}=sqrt{lim f}$ if the limit exists
    – Nosrati
    Nov 24 at 14:49
















$limsqrt{f}=sqrt{lim f}$ if the limit exists
– Nosrati
Nov 24 at 14:49




$limsqrt{f}=sqrt{lim f}$ if the limit exists
– Nosrati
Nov 24 at 14:49










5 Answers
5






active

oldest

votes


















1














By polar coordinates we have that



$$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$



otherwise as an alternative use that



$$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$






share|cite|improve this answer





























    2














    First, your approach is correct. Second, try polar coordinates.






    share|cite|improve this answer





























      1














      $a^2+b^2 ge 2|ab| ge |ab|.$



      $0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$



      $sqrt{ a^2+b^2}.$



      Choose $delta = epsilon$.






      share|cite|improve this answer





























        0














        One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
        $$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
        If we call the height of the triangle $h$, we know that $ba=ch$, hence
        $frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.






        share|cite|improve this answer































          0














          Just another approach



          Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.



          $displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$






          share|cite|improve this answer























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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            By polar coordinates we have that



            $$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$



            otherwise as an alternative use that



            $$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$






            share|cite|improve this answer


























              1














              By polar coordinates we have that



              $$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$



              otherwise as an alternative use that



              $$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$






              share|cite|improve this answer
























                1












                1








                1






                By polar coordinates we have that



                $$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$



                otherwise as an alternative use that



                $$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$






                share|cite|improve this answer












                By polar coordinates we have that



                $$ frac{a^2b^2}{a^2 + b^2}=r^2cos^4thetasin^4theta to 0$$



                otherwise as an alternative use that



                $$0lefrac{a^2b^2}{a^2 + b^2} le frac{(a^2+b^2)^2}{a^2 + b^2}=a^2+b^2 to 0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 at 14:49









                gimusi

                1




                1























                    2














                    First, your approach is correct. Second, try polar coordinates.






                    share|cite|improve this answer


























                      2














                      First, your approach is correct. Second, try polar coordinates.






                      share|cite|improve this answer
























                        2












                        2








                        2






                        First, your approach is correct. Second, try polar coordinates.






                        share|cite|improve this answer












                        First, your approach is correct. Second, try polar coordinates.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 24 at 14:49









                        MisterRiemann

                        5,7691624




                        5,7691624























                            1














                            $a^2+b^2 ge 2|ab| ge |ab|.$



                            $0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$



                            $sqrt{ a^2+b^2}.$



                            Choose $delta = epsilon$.






                            share|cite|improve this answer


























                              1














                              $a^2+b^2 ge 2|ab| ge |ab|.$



                              $0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$



                              $sqrt{ a^2+b^2}.$



                              Choose $delta = epsilon$.






                              share|cite|improve this answer
























                                1












                                1








                                1






                                $a^2+b^2 ge 2|ab| ge |ab|.$



                                $0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$



                                $sqrt{ a^2+b^2}.$



                                Choose $delta = epsilon$.






                                share|cite|improve this answer












                                $a^2+b^2 ge 2|ab| ge |ab|.$



                                $0 le sqrt{|ab| dfrac{|ab|}{a^2+b^2}} le sqrt{ |ab| cdot 1} le$



                                $sqrt{ a^2+b^2}.$



                                Choose $delta = epsilon$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 24 at 17:01









                                Peter Szilas

                                10.6k2720




                                10.6k2720























                                    0














                                    One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
                                    $$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
                                    If we call the height of the triangle $h$, we know that $ba=ch$, hence
                                    $frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.






                                    share|cite|improve this answer




























                                      0














                                      One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
                                      $$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
                                      If we call the height of the triangle $h$, we know that $ba=ch$, hence
                                      $frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.






                                      share|cite|improve this answer


























                                        0












                                        0








                                        0






                                        One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
                                        $$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
                                        If we call the height of the triangle $h$, we know that $ba=ch$, hence
                                        $frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.






                                        share|cite|improve this answer














                                        One may solve the problem geometrically. As the signs of $a$ and $b$ don't matter, let's assume them to be non-negative. Now consider $a$ and $b$ as the legs of a right triangle, then the hypotenuse is $c=sqrt{a^2+b^2}$, hence
                                        $$sqrt{frac{a^2b^2}{a^2+b^2}}=frac{ab}{c}.$$
                                        If we call the height of the triangle $h$, we know that $ba=ch$, hence
                                        $frac{ab}{c}=h$. Now if the legs approach zero, so does the triangle's height.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 24 at 15:20

























                                        answered Nov 24 at 15:09









                                        Michael Hoppe

                                        10.8k31834




                                        10.8k31834























                                            0














                                            Just another approach



                                            Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.



                                            $displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$






                                            share|cite|improve this answer




























                                              0














                                              Just another approach



                                              Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.



                                              $displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$






                                              share|cite|improve this answer


























                                                0












                                                0








                                                0






                                                Just another approach



                                                Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.



                                                $displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$






                                                share|cite|improve this answer














                                                Just another approach



                                                Let $(a, b) to (0, 0)$ along the line $b=am$, $m$ is a constant.



                                                $displaystyle lim_{(a, b) to (0, 0)} sqrt{frac{a^2b^2}{a^2 + b^2}} =displaystyle lim_{(a, am) to (0, 0)} sqrt{frac{a^2(am)^2}{a^2 + (am)^2}}=displaystyle lim_{substack{ato 0\\text{along}: b=am}} sqrt{frac{a^2m^2}{1 + m^2}} =0$







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Nov 24 at 15:37

























                                                answered Nov 24 at 15:31









                                                Yadati Kiran

                                                1,694619




                                                1,694619






























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