If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that $$\lim\limits_{(h,k) \to (0,0)} \dfrac{|f(h,k)|}{|(h,k)|} = 0$$.

For simplicity, I'll be using the euclidean norm for $\Bbb R^n$, $\Bbb R^m$ and $\Bbb R^n\times\Bbb R^m$, and the $\sup$ norm for the target space $\Bbb R^p$. The choice of norms is irrelevant to the result since all norms are equivalent in finite dimension.

There are coefficients $a_{ij;\,l}\in\Bbb R$, where $(i,j)$ ranges over $\lbrace 1,\dots,n\rbrace\times\lbrace 1,\dots,m\rbrace$ and $k\in\lbrace 1,\dots,p\rbrace$ such that for all $h=(h_1,\dots,h_n)\in\Bbb R^n,k=(k_1,\dots,k_m)\in\Bbb R^m$, the $l$-th coordinate of $f(h,k)$ is equal to 
$$\left[f(h,k)\right]_l=\sum_{i,j}a_{ij;\,l}h_ik_j$$
If you let $A$ be the maximum of the absolute values $|a_{ij;\,l}|$, then for all $l$,
$$|\left[f(h,k)\right]_l|\leq A\sum_{i,j}|h_ik_j|\leq \frac12A\sum_{i,j}h_i^2+k_j^2=\frac12A(m|h|^2+n|k|^2)\leq\frac12A(m+n)|(h,k)|^2$$
Thus, for all $(h,k)\neq(0,0)$
$$\frac{|f(h,k)|}{|(h,k)|}\leq C|(h,k)|$$
for $C=\frac12A(m+n)$. The right hand side tends to zero as $(h,k)$ tends to zero.
