How i can solve this limit? I Try substituting $x=r\cos\theta$ , $y=r\sin\theta$, but it has not worked. $$\lim_{(x,y) \to (0,0)}{\frac{\sin(x^2+y^2)}{x^2+y^2}}$$

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    $\begingroup$ $\lim_{t \to 0} \sin(t) / t = 1$. $\endgroup$ – user296602 Aug 25 '16 at 19:26
  • $\begingroup$ This must be $1$ because you can change $x^2+y^2=z$ and you have $\frac{\sin z}{z}$ $\endgroup$ – Piquito Aug 25 '16 at 19:39

Use polar transformation: $x^2+y^2=r^2$ then the limit becomes $$ \lim_{r\rightarrow 0}\frac{\sin(r^2)}{r^2}=1 $$


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