Let $f:\mathbb R^n\to \mathbb R$ be a continuous function with the property that there exits two sequence $(b_n)$ and $(c_n)\in \mathbb R^n$ such that Let $f:\mathbb R^n\to \mathbb R$ be a continuous function with the property that there exits two sequences $(b_n)$ and $(c_n)\in \mathbb R^n$ such that
$f(b_n)\to \infty$ and $f(c_n)\to -\infty $
Now I want to prove that $\lim\limits_{n\to \infty}||b_n||\to\infty$ and $\lim\limits_{n\to \infty}||c_n||\to\infty$ 
and  I want to show that $k>1$ there exists some $(a_n)\in\mathbb R^k$ such that $f(a_n)=0$ for all $n\in\mathbb N$ and $||a_n||\to \infty$
My work for the  $\lim\limits_{n\to \infty}||b_n||\to\infty$ and $\lim\limits_{n\to \infty}||c_n||\to\infty$ 
Assume not so  $||b_n||\to b\in \mathbb R$ and similar for $c_n$, $||c_n||\to c\in \mathbb R$
so since $||.||$ is a continuous function so $b_n\to \vec b\in\mathbb R^n$ and $c_n\to \vec c\in\mathbb R^n$ so since $f$ is continuous.
$$f(b_n)\to f(\vec b)\in\mathbb R$$ so it cannot be infinity. How to do it properly and can you give me a hint about second question I am stuck.$n>1$ there exists some $(a_n)\in\mathbb R^n$ such that $f(a_n)=0$ for all $n\in\mathbb N$ and $||a_n||\to \infty$

Suppose that $lim_n\|b_n\|$ is not $\infty$, there exists $C$ such that for every $n$, there exists $p_n>n$ with $\|b_n\|<C$.  Let $B(0,C)$ be the closed ball of radius $f(B,0,C))$, since $f$ is continuous, $f(B(0,C))$ is compact  and bounded, and contained in $[-P,P]$. Since $lim_nf(b_n)=\infty$,  there exists $n_P$ such that $n>n_R$ implies that $|f(b_n)|>P$. Contradiction.
