I've been trying to check the continuity of the following function:
$$ f(x,y) = \begin{cases} \frac {(x-1)(y-4)^2}{(x-1)^2+\sin(y-4)} & \quad \text{(x,y) } \ne \text{ (1,4)}\\ 0 & \quad \text{(x,y) } = \text{ (1,4)} \end{cases} $$
I've tried calculating the following $lim$ , as $t = x-1$ , and $ z = y-4 $ :
$$ \lim_{{(t,z)\to(0,0)}}{ \frac {tz^2}{t^2+\sin(z)}} $$
I've tried choosing different paths: $ t=z$ and $t=z^2$ but both gave me the same result - $0$ .
I'm not sure how to prove or disprove that the limit is $0$ .
I'd appreciate your help, thanks!
Note that for $z=\arcsin (t^5-t^2)\to 0$
$$\frac {tz^2}{t^2+\sin(z)}=\frac {t(\arcsin (t^5-t^2))^2}{t^2+t^5-t^2}\to 1$$