# Calculate the following limits, if they exist, by using a combination of polar c

Calculate the following limits, if they exist, by using a combination of polar coordinates and de L’Hopital rule.
$\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{1-\mathrm{cos}\left({x}^{2}+{y}^{2}\right)}{{x}^{2}+{y}^{2}}$
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Given: The function is $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{1-\mathrm{cos}\left({x}^{2}+{y}^{2}\right)}{{x}^{2}+{y}^{2}}$
Consider $x=r\mathrm{cos}\theta$, and $y=r\mathrm{sin}\theta$.
The expression becomes $\underset{r\to 0}{lim}\frac{1-\mathrm{cos}\left({r}^{2}\right)}{{r}^{2}}$
Conclusion:
Apply L’Hopital rule as follows:
$\underset{r\to 0}{lim}\frac{1-\mathrm{cos}\left({r}^{2}\right)}{{r}^{2}}=\underset{r\to 0}{lim}\frac{-\left(-\mathrm{sin}\left({r}^{2}\right)\cdot 2r\right)}{2r}$
$=\underset{r\to 0}{lim}\mathrm{sin}\left({r}^{2}\right)$
$=\mathrm{sin}\left(0\right)$
$=0$
Hence, $\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{1-\mathrm{cos}\left({x}^{2}+{y}^{2}\right)}{{x}^{2}+{y}^{2}}=0$

Jeffrey Jordon