Determining the limit of $sqrt{x^2+y^2} ln(x^2+y^2)$ as $(x,y)$ approaches $(0,0)$












3












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I want to evaluate the limit of $sqrt{x^2+y^2}ln(x^2+y^2)$ as $(x,y)$ approaches $(0,0)$. Using polar coordinates, $x=r cos theta$, $y= r sin theta$, I obtain $sqrt{x^2+y^2}ln(x^2+y^2)=2rln(r)$ but I can not understand why this would be more simple than the original expression, or how I can proceed with this problem.






limits multivariable-calculus







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  • $begingroup$
    Tip: You can use $LaTeX$ in titles.
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    – Mohammad Zuhair Khan
    Jan 27 at 8:08
















3












$begingroup$


I want to evaluate the limit of $sqrt{x^2+y^2}ln(x^2+y^2)$ as $(x,y)$ approaches $(0,0)$. Using polar coordinates, $x=r cos theta$, $y= r sin theta$, I obtain $sqrt{x^2+y^2}ln(x^2+y^2)=2rln(r)$ but I can not understand why this would be more simple than the original expression, or how I can proceed with this problem.






limits multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Tip: You can use $LaTeX$ in titles.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 27 at 8:08














3












3








3





$begingroup$


I want to evaluate the limit of $sqrt{x^2+y^2}ln(x^2+y^2)$ as $(x,y)$ approaches $(0,0)$. Using polar coordinates, $x=r cos theta$, $y= r sin theta$, I obtain $sqrt{x^2+y^2}ln(x^2+y^2)=2rln(r)$ but I can not understand why this would be more simple than the original expression, or how I can proceed with this problem.






limits multivariable-calculus







share|cite|improve this question











$endgroup$




I want to evaluate the limit of $sqrt{x^2+y^2}ln(x^2+y^2)$ as $(x,y)$ approaches $(0,0)$. Using polar coordinates, $x=r cos theta$, $y= r sin theta$, I obtain $sqrt{x^2+y^2}ln(x^2+y^2)=2rln(r)$ but I can not understand why this would be more simple than the original expression, or how I can proceed with this problem.






limits multivariable-calculus




limits multivariable-calculus






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edited Jan 27 at 8:13









Robert Z

101k1070143




101k1070143










asked Jan 27 at 8:05









JohannaJohanna

858




858












  • $begingroup$
    Tip: You can use $LaTeX$ in titles.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 27 at 8:08


















  • $begingroup$
    Tip: You can use $LaTeX$ in titles.
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 27 at 8:08
















$begingroup$
Tip: You can use $LaTeX$ in titles.
$endgroup$
– Mohammad Zuhair Khan
Jan 27 at 8:08




$begingroup$
Tip: You can use $LaTeX$ in titles.
$endgroup$
– Mohammad Zuhair Khan
Jan 27 at 8:08










1 Answer
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Your approach with polar coordinates is correct. The limit $lim_{rto 0^+}2rln(r)$ is simpler than the given one because it is a one-variable limit. One way to evaluate it is to write $2rln(r)$ as $frac{2ln(r)}{1/r}$ and then apply L'Hopital. Can you take it from here?






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  • 1




    $begingroup$
    Yes, the limit seems to exist and is 0.
    $endgroup$
    – Johanna
    Jan 27 at 8:25






  • 1




    $begingroup$
    Yes, you are correct!
    $endgroup$
    – Robert Z
    Jan 27 at 8:37











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Your approach with polar coordinates is correct. The limit $lim_{rto 0^+}2rln(r)$ is simpler than the given one because it is a one-variable limit. One way to evaluate it is to write $2rln(r)$ as $frac{2ln(r)}{1/r}$ and then apply L'Hopital. Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yes, the limit seems to exist and is 0.
    $endgroup$
    – Johanna
    Jan 27 at 8:25






  • 1




    $begingroup$
    Yes, you are correct!
    $endgroup$
    – Robert Z
    Jan 27 at 8:37
















3












$begingroup$

Your approach with polar coordinates is correct. The limit $lim_{rto 0^+}2rln(r)$ is simpler than the given one because it is a one-variable limit. One way to evaluate it is to write $2rln(r)$ as $frac{2ln(r)}{1/r}$ and then apply L'Hopital. Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Yes, the limit seems to exist and is 0.
    $endgroup$
    – Johanna
    Jan 27 at 8:25






  • 1




    $begingroup$
    Yes, you are correct!
    $endgroup$
    – Robert Z
    Jan 27 at 8:37














3












3








3





$begingroup$

Your approach with polar coordinates is correct. The limit $lim_{rto 0^+}2rln(r)$ is simpler than the given one because it is a one-variable limit. One way to evaluate it is to write $2rln(r)$ as $frac{2ln(r)}{1/r}$ and then apply L'Hopital. Can you take it from here?






share|cite|improve this answer









$endgroup$



Your approach with polar coordinates is correct. The limit $lim_{rto 0^+}2rln(r)$ is simpler than the given one because it is a one-variable limit. One way to evaluate it is to write $2rln(r)$ as $frac{2ln(r)}{1/r}$ and then apply L'Hopital. Can you take it from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 8:10









Robert ZRobert Z

101k1070143




101k1070143








  • 1




    $begingroup$
    Yes, the limit seems to exist and is 0.
    $endgroup$
    – Johanna
    Jan 27 at 8:25






  • 1




    $begingroup$
    Yes, you are correct!
    $endgroup$
    – Robert Z
    Jan 27 at 8:37














  • 1




    $begingroup$
    Yes, the limit seems to exist and is 0.
    $endgroup$
    – Johanna
    Jan 27 at 8:25






  • 1




    $begingroup$
    Yes, you are correct!
    $endgroup$
    – Robert Z
    Jan 27 at 8:37








1




1




$begingroup$
Yes, the limit seems to exist and is 0.
$endgroup$
– Johanna
Jan 27 at 8:25




$begingroup$
Yes, the limit seems to exist and is 0.
$endgroup$
– Johanna
Jan 27 at 8:25




1




1




$begingroup$
Yes, you are correct!
$endgroup$
– Robert Z
Jan 27 at 8:37




$begingroup$
Yes, you are correct!
$endgroup$
– Robert Z
Jan 27 at 8:37


















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