I want to show that the function $$ f: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto \begin{cases} x^2(x-1)(y-1)\sin(xy), & (x,y) \in [0,1]^2 \\ 0, & \text{elsewhere.} \end{cases} $$ is continuous.
Obviously, both pieces are continuous because they are composed of elementary continuous functions. Now I only need to show that the transition between both pieces is continuous and don't know how to notate it properly. My Idea was \begin{align*} \lim_{\|x,y\|_{\infty} \nearrow 1} f|_{[0,1]} & = \lim_{\max(x,y) \nearrow 1} x^2(x - 1)(y - 1)\sin(xy) \\ & = \begin{cases} \lim_{x \nearrow 1} x^2(x - 1)(y - 1)\sin(xy) \\ \lim_{y \nearrow 1} x^2(x - 1)(y - 1)\sin(xy) \end{cases} = \begin{cases} 1^2(1 - 1)(y - 1)\sin(y) \\ x^2(x - 1)(1 - 1)\sin(x) \end{cases} = 0 \end{align*} I've never seen it done that way but didn't have a better idea.
Second Attempt For all $a \in [0,1]$ we have \begin{equation*} \begin{cases} \lim\limits_{(x,y)\to(0,a)} f(x,y) = 0^2(0 - 1)(a - 1)\sin(0) = 0 = f(0,a), \\ \lim\limits_{(x,y)\to(1,a)} f(x,y) = 1^2(1 - 1)(a - 1)\sin(y) = 0 = f(1,a). \\ \lim\limits_{(x,y)\to(a,0)} f(x,y) = a^2(a - 1)(0 - 1)\sin(0) = 0 = f(a,0), \\ \lim\limits_{(x,y)\to(a,1)} f(x,1) = a^2(a - 1)(1 - 1)\sin(x) = 0 = f(a,1) \end{cases} \end{equation*}
To prove continuity we need to show that for $0\le a\le 1$