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Prove by definition $\epsilon-\delta$ $$\lim_{(x,y)\rightarrow(0,0)}\frac{1-\cos(xy)}{(xy)^2}=\frac{1}{2}$$

My work:

Let $\epsilon >0$, and $\delta =...$
If $\sqrt{x^2+y^2}<\delta$ then $|\frac{1-cos(xy)}{(xy)^2}-\frac{1}{2}|=|\frac{2-2cos(xy)-(xy)^2}{2(xy)^2}|=\frac{|2-2cos(xy)-(xy)^2|}{2(xy)^2}\leq\frac{|2-2-(xy)^2|}{2(xy)^2}=\frac{|-(xy)^2|}{2(xy)^2}=\frac{(xy)^2}{2(xy)^2}=\frac{1}{2}$

In this step i'm stuck. Can someone help me?

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    $\begingroup$ Careful with that inequality. While it's true that $2\cos(xy) \le 2$, the presence of negatives and absolute values seriously complicate things! $\endgroup$ – Theo Bendit Nov 8 '17 at 1:29
  • $\begingroup$ That limit is a little complicated. @TheoBendit Thanks for the comment. $\endgroup$ – Bvss12 Nov 8 '17 at 1:33
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    $\begingroup$ Use $|\sin t|<|t|$. $\endgroup$ – Nosrati Nov 8 '17 at 1:51
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With $|\sin t|\leq|t|$ or $-t^2\leq-\sin^2t$ we write $$2-2\cos t-t^2\leq2-2\cos t-\sin^2t=4\sin^2\dfrac12t-4\sin^2\dfrac12t\cos^2\dfrac12t=4\sin^4\dfrac12t$$ then $$\Big|\dfrac{1-\cos t}{t^2}-\dfrac12\Big|=\Big|\dfrac{2-2\cos t-t^2}{2t^2}\Big|\leq\Big|\dfrac{4\sin^4\dfrac12t}{2t^2}\Big|\leq\Big|\dfrac{\dfrac14t^4}{2t^2}\Big|=\dfrac18t^2$$ also $|xy|\leq\dfrac12(x^2+y^2)$.

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