glasskerfu
2021-10-22
Answered

Calculate the following limits, if they exist, by using a combination of polar coordinates and de L’Hopital rule.

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

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l1koV

Answered 2021-10-23
Author has **100** answers

changing into polar coordinates

$x=r\mathrm{cos}\theta$

$y=r\mathrm{sin}\theta$

${x}^{2}+{y}^{2}={r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta$

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

$={r}^{2}({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1)$

when$(x,y)\to (0,0)$

$r\to 0$

Now$\underset{(x,y)\to (0,0)}{lim}\frac{\mathrm{arctan}({x}^{2}+{y}^{2})}{{x}^{2}+{y}^{2}}$

$=\underset{r\to 0}{lim}\frac{\mathrm{arctan}\left({r}^{2}\right)}{{r}^{2}}$

$=\underset{r\to 0}{lim}\frac{\frac{1}{1+{\left({r}^{2}\right)}^{2}}\cdot 2r}{2r}$ (applying L'Hopital rule)

$=\underset{r\to 0}{lim}\frac{1}{1+{r}^{4}}$

$=1$

when

Now

Jeffrey Jordon

Answered 2022-06-26
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