Question

Find the solution of limit displaystylelim_{{{left({x},{y}right)}rightarrow{0},{0}}}frac{{sqrt{{{x}^{2}+{y}^{2}}}}}{{{x}^{2}+{y}^{2}}} by using the polar coordinates system.

Find the solution of limit \(\displaystyle\lim_{{{\left({x},{y}\right)}\rightarrow{0},{0}}}\frac{{\sqrt{{{x}^{2}+{y}^{2}}}}}{{{x}^{2}+{y}^{2}}}\) by using the polar coordinates system.

Answers (1)

2021-01-18
\(\displaystyle\lim_{{{\left({x},{y}\right)}\rightarrow{0},{0}}}{\left(\sqrt{{{x}^{2}+{y}^{2}}}{\left({x}^{2}+{y}^{2}\right)}\right)}=\lim_{{{\left({x},{y}\right)}\rightarrow{0},{0}}} f{{\left({x},{y}\right)}}\),
where \(\displaystyle f{{\left({x},{y}\right)}}={\left(\sqrt{{{x}^{2}+{y}^{2}}}\right)}.\text{Using polar form}\ {x}={r} \cos{{0}},{y}={r} \sin{{0}}\) we have
\(\displaystyle f{{\left({r},{0}\right)}}=\frac{r}{{{r}^{2}{\left({{\cos}^{2}{0}}+{{\sin}^{2}{0}}\right)}}}=\frac{r}{{r}^{2}}\)
Now
\(\displaystyle\lim_{{{r}\to{0},{0}\to{0}}} f{{\left({r},{0}\right)}}=\lim_{{{r}\to{0}}}\frac{r}{{r}^{2}}=\infty\)
Therefore
\(\displaystyle\lim_{{{\left({x},{y}\right)}\rightarrow{0},{0}}}\frac{{\sqrt{{x}}^{2}+{y}^{2}}}{{{x}^{2}+{y}^{2}}}=\infty\)
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