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 $t=x^2+y^2$.)

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Here’s a point of view I like. The fact that $$\lim_{t\to0}\frac{\sin t}{t}=1$$ means precisely that the function $\phi:\mathbb R\to\mathbb R$ defined by $$\phi(t)=\begin{cases}\frac{\sin t}t;&t\neq0,\\1;&t=0,\end{cases}$$ is continuous. Therefore, we have $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{(x,y)\to(0,0)}\phi(x^2+y^2)=\phi(0)=1.$$ The first equality holds because the definition of limit doesn’t involve the value of the function at $(0,0)$ and the second equality holds because $\phi(x^2+y^2)$ is continuous (since it is the composition of two continuous functions).

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 $.