Continuity of multivariable function limit at $0$

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Let $f:mathbbR^2 to mathbbR,$ $ f(x)=begincasesfracx_1^2cdot x_2x_1^2+x_2^2 & , x ne 0 \ 0 &, x = 0. endcases$



Show that f is continuous at $0$.



I tried this way ,I am not sure if it is true.



Suppose $left|xright|<a$ i.e. $(x_1^2+x_2^2)^frac12<a.$



Now,
beginalign
|f(x)-f(0)|&=left|fracx_1^2cdot x_2x_1^2+x_2^2right|\
& lefrac\
&<|x_2|\
&<a.
endalign
Then by definition of continuity $f$ is continuous at $0$.




limits multivariable-calculus




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  • welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
    – Siong Thye Goh
    Jul 21 at 2:40










  • Sound perfect.....
    – Mostafa Ayaz
    Jul 21 at 18:49














up vote
-2
down vote

favorite
1












Let $f:mathbbR^2 to mathbbR,$ $ f(x)=begincasesfracx_1^2cdot x_2x_1^2+x_2^2 & , x ne 0 \ 0 &, x = 0. endcases$



Show that f is continuous at $0$.



I tried this way ,I am not sure if it is true.



Suppose $left|xright|<a$ i.e. $(x_1^2+x_2^2)^frac12<a.$



Now,
beginalign
|f(x)-f(0)|&=left|fracx_1^2cdot x_2x_1^2+x_2^2right|\
& lefrac\
&<|x_2|\
&<a.
endalign
Then by definition of continuity $f$ is continuous at $0$.




limits multivariable-calculus




share|cite|improve this question





















  • welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
    – Siong Thye Goh
    Jul 21 at 2:40










  • Sound perfect.....
    – Mostafa Ayaz
    Jul 21 at 18:49












up vote
-2
down vote

favorite
1









up vote
-2
down vote

favorite
1






1





Let $f:mathbbR^2 to mathbbR,$ $ f(x)=begincasesfracx_1^2cdot x_2x_1^2+x_2^2 & , x ne 0 \ 0 &, x = 0. endcases$



Show that f is continuous at $0$.



I tried this way ,I am not sure if it is true.



Suppose $left|xright|<a$ i.e. $(x_1^2+x_2^2)^frac12<a.$



Now,
beginalign
|f(x)-f(0)|&=left|fracx_1^2cdot x_2x_1^2+x_2^2right|\
& lefrac\
&<|x_2|\
&<a.
endalign
Then by definition of continuity $f$ is continuous at $0$.




limits multivariable-calculus




share|cite|improve this question













Let $f:mathbbR^2 to mathbbR,$ $ f(x)=begincasesfracx_1^2cdot x_2x_1^2+x_2^2 & , x ne 0 \ 0 &, x = 0. endcases$



Show that f is continuous at $0$.



I tried this way ,I am not sure if it is true.



Suppose $left|xright|<a$ i.e. $(x_1^2+x_2^2)^frac12<a.$



Now,
beginalign
|f(x)-f(0)|&=left|fracx_1^2cdot x_2x_1^2+x_2^2right|\
& lefrac\
&<|x_2|\
&<a.
endalign
Then by definition of continuity $f$ is continuous at $0$.





limits multivariable-calculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 2:56









Siong Thye Goh

77.6k134795




77.6k134795









asked Jul 21 at 2:22









S.Chauhan

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62











  • welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
    – Siong Thye Goh
    Jul 21 at 2:40










  • Sound perfect.....
    – Mostafa Ayaz
    Jul 21 at 18:49
















  • welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
    – Siong Thye Goh
    Jul 21 at 2:40










  • Sound perfect.....
    – Mostafa Ayaz
    Jul 21 at 18:49















welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
– Siong Thye Goh
Jul 21 at 2:40




welcome to MSE. I'm glad that you included your attempt. I believed you are downvoted because your post failed to display your attempt correctly. Here's a mathjax tutorial to help you learn how to typeset on this site.
– Siong Thye Goh
Jul 21 at 2:40












Sound perfect.....
– Mostafa Ayaz
Jul 21 at 18:49




Sound perfect.....
– Mostafa Ayaz
Jul 21 at 18:49










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I think it should be $$fracx_1^2+x_2^2le |x_2|.$$



Since the inequality might hold if $x_2=0$, other than that, it seems fine.






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    up vote
    0
    down vote













    I think it should be $$fracx_1^2+x_2^2le |x_2|.$$



    Since the inequality might hold if $x_2=0$, other than that, it seems fine.






    share|cite|improve this answer

























      up vote
      0
      down vote













      I think it should be $$fracx_1^2+x_2^2le |x_2|.$$



      Since the inequality might hold if $x_2=0$, other than that, it seems fine.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I think it should be $$fracx_1^2+x_2^2le |x_2|.$$



        Since the inequality might hold if $x_2=0$, other than that, it seems fine.






        share|cite|improve this answer













        I think it should be $$fracx_1^2+x_2^2le |x_2|.$$



        Since the inequality might hold if $x_2=0$, other than that, it seems fine.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 21 at 2:33









        Siong Thye Goh

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