<s>[INST] Why $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim\limits_{t\to 0}\frac{\sin t}{t}$? Why $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{t\to 0}\frac{\sin t}{t}$( and hence equals to $1$)?
Any rigorous reason? (i.e. not just say by letting $t=x^2+y^2$.)
 [/INST] Using $x = t \cos \alpha $ and $y = t \sin \alpha$, then:
$$ \frac{ \sin (x^2 + y^2 ) }{x^2+y^2} =  \frac{ \sin (t^2(\sin^2 \alpha + \cos^2 \alpha)) }{t^2(\sin^2 \alpha + \cos^2 \alpha)} = \frac{ \sin (t^2) }{t^2}$$
.
Note as $x,y \to 0,0$, then $t \to 0 $.
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