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How can I calculate this limit $$\lim_{(x,y)\to(0,0)} \dfrac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$$ at the origin?

I tried to use the substitution $x ^2 + y^2=t$ but how can I evaluate the value of $xy$? I even tried to use polar coordinates but to avail.

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For $xy$, note that $(x-y)^2\ge 0$, so $|xy|\le \frac{1}{2}(x^2+y^2)$. – André Nicolas Mar 19 at 7:31

Outline: Note that $(x-y)^2\ge 0$, so $|xy|\le \frac{1}{2}(x^2+y^2)$. One can also get this from polar coordinates, for $|xy|=r^2|\cos\theta\sin\theta|=\frac{1}{2}r^2|\sin(2\theta)|\le \frac{r^2}{2}$.

Now you can comfortably let $t=x^2+y^2$. You will need to look at the behaviour of $1-\cos t$ near $0$. This can be taken care of mechanically by using L'Hospital's Rule. Or else one can use the beginning of the Maclaurin series for $\cos t$. Or else one can note that $1-\cos t=\frac{1-\cos^2 t}{1+\cos t}=\frac{\sin^2 t}{1+\cos t}$.

The conclusion will be that the limit is $0$.

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Using polar coordinates $x=r\cos\theta,y=r\sin\theta$ you get $$ \begin{align*}0&\leq \lim_{(x,y)\to(0,0)} \left|\dfrac{xy(1-\cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}\right|=\lim_{r\to 0} \left|\dfrac{r^2\cos\theta\sin\theta(1-\cos(r^2))}{(r^2)^{5/2}}\right|\\&=\lim_{r\to 0}\left|\dfrac{1-\cos(r^2)}{r^3}\right|\cdot\left|(\cos\theta\sin\theta)\right|\\ &\leq \lim_{r\to 0}\left|\dfrac{\sin(r^2)\cdot 2r}{3r^2}\right|=\left|\lim_{r\to 0}\dfrac{2}{3}\dfrac{\sin(r^2)}{r}\right|=\left|\lim_{r\to 0}\dfrac{2}{3}\dfrac{2r\cos(r^2)}{1}\right|=|0|=0.\end{align*}$$

So, $$\lim_{(x,y)\to(0,0)} \dfrac{xy(1-\cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}=0$$

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I decided to correct it instead. – Wore Mar 19 at 8:15
thanks to both of you – Rayees Ahmad Mar 19 at 8:30

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