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 $||a_n||\to \infty$
 A: 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.
A: Every continuous function reaches its maximum (resp. minimum) on every compact subset.
So, if $b_n$ is bounded in norm then, then we must have also $f(b_n)$ bounded in norm.
For the second part, suppose we have two sequences $\left( a_{n}\right)
_{n\geq 0}$ and $\left( b_{n}\right) _{n\geq 0}$ such that $\lim \left\Vert
a_{n}\right\Vert =\lim \left\Vert a_{n}\right\Vert =+\infty $ and $\lim
f\left( a_{n}\right) =-\lim f$ $\left( b_{n}\right) =+\infty $. Without loss
of generality we can take $f\left( a_{n}\right) $ and $-f\left( b_{n}\right) 
$ positive for every $n\geq 0$.
It is possible to take a sequence of curves $\left( S_{n}\right) _{n\geq 0}\ 
$such that $S_{n}$ joins $a_{n}$ to $b_{n}$ and for every $x\in S_{n}$ we
have,$\left\Vert x\right\Vert \geq \min (\left\Vert a_{n}\right\Vert
,\left\Vert b_{n}\right\Vert )$.
Since $f$ is continuous on $S_{n}$,  $f\left( a_{n}\right) >0$ and $f\left(
b_{n}\right) <0$ for every $n\geq 0$, we have a point $c_{n}\in S_{n}$ such
that $f(c_{n})=0$ and $\left\Vert c_{n}\right\Vert \geq \min (\left\Vert
a_{n}\right\Vert ,\left\Vert b_{n}\right\Vert )$. 
Now it is easy to check that $\lim \left\Vert c_{n}\right\Vert =\infty $.
A: The other answers already give good arguments for the $(b_n)$ and $(c_n)$ part.
I will answer the $(a_n)$ part.
Let $\Gamma_n$ a path between $b_n$ and $c_n$ such that its smallest point is greater than $\frac 1 2 \min (b_n,c_n)$ in norm. Just pick a path between $b_n$ and $c_n$ outside the ball centered at $0$ with radius $\frac 1 2 \min (b_n,c_n)$ (which is always possible for $k \ge 2$ since we still have a connected space after removing this ball). Define $g_n$ the restriction of $f$ over this path, by the intermediate value theorem over $g_n$, we can find $a_n \in \Gamma_n$ such that $f(a_n) =0$.
By construction $ \|a_n\| \ge \frac 1 2 \min (b_n,c_n)$ and we conclude.
