alesterp

2021-02-05

Use polar coordinates to find the limit. [Hint: Let $x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}$ , and note that (x, y) (0, 0) implies r 0.] $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}$

stuth1

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

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