Prove that $\lim_{t\to t_0}[f(t) \times g(t)]=u \times v$ Let $f(t)=(f_1(t),f_2(t),f_3(t))$, $g(t)=(g_1(t),g_2(t),g_3(t)).$
$$\lim_{t\to t_0}f(t)=u; \lim_{t\to t_0}g(t)=v.$$
Prove: $$\lim_{t\to t_0}[f(t) \times g(t)]=u \times v$$.
Thanks ahead:)
 A: The function
$$(u,v)\longmapsto u\times v=(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1)$$
is obviously continuous, so
$$
\lim_{t\to 0}[f(t)\times g(t)]=\lim_{t\to 0}f(t)\times\lim_{t\to 0}g(t)=u\times v.
$$
A: prove:
$$\lim_{t\to t_0}[f(t)\times g(t)]\\=\lim_{t\to t_0}\left|\matrix{\overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\f_1(t) & f_2(t) & f_3(t)\\g_1(t) & g_2(t) & g_3(t)}\right|\\
=\lim_{t\to t_0}[(f_2(t)g_3(t)-f_3(t)g_2(t))\overrightarrow{i}+(f_3(t)g_1(t)-f_1(t)g_3(t))\overrightarrow{j}+(f_1(t)g_2(t)-f_2(t)g_1(t))\overrightarrow{k}]\\
=\lim_{t\to t_0}(f_2(t)g_3(t)-f_3(t)g_2(t))\overrightarrow{i}+\lim_{t\to t_0}(f_3(t)g_1(t)-f_1(t)g_3(t))\overrightarrow{j}+\lim_{t\to t_0}(f_1(t)g_2(t)-f_2(t)g_1(t))\overrightarrow{k}$$
if $\lim_{t\to t_0}f(t),\lim_{t\to t_0}g(t)$ exist, recall that: $$\lim_{t\to t_0}(f(t)\cdot g(t))=\lim_{t\to t_0}f(t)\cdot\lim_{t\to t_0}g(t)$$
so:
$$\lim_{t\to t_0}(f_2(t)g_3(t)-f_3(t)g_2(t))\overrightarrow{i}+\lim_{t\to t_0}(f_3(t)g_1(t)-f_1(t)g_3(t))\overrightarrow{j}+\lim_{t\to t_0}(f_1(t)g_2(t)-f_2(t)g_1(t))\overrightarrow{k}\\=(u_2v_3-u_3v_2)\overrightarrow{i}+(u_3v_1-u_1v_3)\overrightarrow{j}+(u_1v_2-u_2v_1)\overrightarrow{k}\\=\left|\matrix{\overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3}\right|\\=u\times v$$
