alesterp

2021-02-05

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

stuth1

Skilled2021-02-06Added 97 answers

Given the multivariable limit function:

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}$

For polar coordinate system:

$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=\underset{r\to 0}{lim}\frac{{r}^{2}{\mathrm{cos}}^{2}\theta -{r}^{2}{\mathrm{sin}}^{2}\theta}{\sqrt{{r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta}}$

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=\underset{r\to 0}{lim}\frac{{r}^{2}({\mathrm{cos}}^{2}\theta -{r}^{2}{\mathrm{sin}}^{2}\theta )}{\sqrt{{r}^{2}({\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta )}}$

We know the trigonometry identities

$({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta )=\mathrm{cos}2\theta$

$\left({\mathrm{cos}}^{2}\theta 1{\mathrm{sin}}^{2}\theta \right)=1$

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=\underset{r\to 0}{lim}\frac{{r}^{2}\mathrm{cos}2\theta}{r}$

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=\underset{r\to 0}{lim}r\mathrm{cos}2\theta$

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}=0$

Hence the limit of the given function is zero.

For polar coordinate system:

We know the trigonometry identities

Hence the limit of the given function is zero.

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1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$

2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?