Use polar coordinates to find the limit. [Hint: Let x = r cos and y = r sin , and note that (x, y) (0, 0) implies r 0.] lim_((x,y)->(0,0)) (x^2-y^2)/sqrt(x^2+y^2)

Use polar coordinates to find the limit. [Hint: Let x = r cos and y = r sin , and note that (x, y) (0, 0) implies r 0.] lim_((x,y)->(0,0)) (x^2-y^2)/sqrt(x^2+y^2)

Question
Multivariable functions
asked 2021-02-05
Use polar coordinates to find the limit. [Hint: Let \(\displaystyle{x}={r}{\cos{{\quad\text{and}\quad}}}{y}={r}{\sin{}}\) , and note that (x, y) (0, 0) implies r 0.] \(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}\)

Answers (1)

2021-02-06
Given the multivariable limit function:
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}\)
For polar coordinate system:
\(\displaystyle{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{{\cos}^{{2}}\theta}-{r}^{{2}}{{\sin}^{{2}}\theta}}}{\sqrt{{{r}^{{2}}{{\cos}^{{2}}\theta}+{r}^{{2}}{{\sin}^{{2}}\theta}}}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{\left({{\cos}^{{2}}\theta}-{r}^{{2}}{{\sin}^{{2}}\theta}\right)}}}{\sqrt{{{r}^{{2}}{\left({{\cos}^{{2}}\theta}+{r}^{{2}}{{\sin}^{{2}}\theta}\right)}}}}\)
We know the trigonometry identities
\(\displaystyle{\left({{\cos}^{{2}}\theta}-{{\sin}^{{2}}\theta}\right)}={\cos{{2}}}\theta\)
\(\displaystyle{\left({{\cos}^{{2}}\theta}{1}{{\sin}^{{2}}\theta}\right)}={1}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}\frac{{{r}^{{2}}{\cos{{2}}}\theta}}{{r}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}=\lim_{{{r}\to{0}}}{r}{\cos{{2}}}\theta\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}={0}\)
Hence the limit of the given function is zero.
0

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