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$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$$

I tried to bound it with $\frac{\sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = r\cos\theta$ and $y = r\sin\theta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?

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Use $|\sin t|\leq |t|$ then $$\Big|\frac{\sin(x^4+y^4)}{x^2+y^2}\Big|\leq\frac{x^4+y^4}{x^2+y^2}\leq\frac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$

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