Let be $P:\mathbb{R}\to\mathbb{R}$ where $P(x):=\sum\limits_{k=0}^na_kx^k$ and $a_n\neq0$, $n$ is odd. Show that $P(x)$ attains every $y\in\mathbb{R}$, i.e. for each $y\in\mathbb{R}$ there exists at least one $x\in\mathbb{R}$ such that $P(x)=y$.
My approach:
Let's assume that $a_n>0$.
We know that $$\lim\limits_{x\to\infty}P(x)=\lim\limits_{x\to\infty}\left(x^n\cdot\left(\sum\limits_{k=0}^{n}a_kx^{k-n}\right)\right)=\underset{=\infty}{\underbrace{\lim\limits_{x\to\infty}x^n}}\cdot \underset{=a_n}{\underbrace{\lim\limits_{x\to\infty}\left(\sum\limits_{k=0}^{n}a_kx^{k-n}\right)}}=\infty$$ and $$\lim\limits_{x\to-\infty}P(x)=\lim\limits_{x\to-\infty}\left(x^n\cdot\left(\sum\limits_{k=0}^{n}a_kx^{k-n}\right)\right)=\underset{=-\infty\text{ as n is odd}}{\underbrace{\lim\limits_{x\to-\infty}x^n}}\cdot \underset{=a_n}{\underbrace{\lim\limits_{x\to-\infty}\left(\sum\limits_{k=0}^{n}a_kx^{k-n}\right)}}=-\infty.$$
So by definition of the "$\infty$-limit" for each $c_1\in\mathbb{R}$ there exists a sufficiently large $\delta_1>0$ such that for all $x>\delta$ we have $P(x)>c$.
Again by definition of the "$-\infty$-limit" for each $c_0\in\mathbb{R}$ there exists a sufficiently small $\delta_0<0$ such that for all $x<\delta$ we have $P(x)<c$. We set $c_1:=y+1$ and $c_0:=y-1$.
Then we know that there exist $x_0$ and $x_1$ such that $P(x_0)<y-1<y<y+1<P(x_1)$.
As $P(x)$ is continuous we can apply the intermediate value theorem which guarantees the existence of a $x$ with $P(x)=y$. If $a_n<0$ then we can basically use the same argument with flipped signs. As $y$ was chosen arbitrarily $P(x)$ attains every $y\in \mathbb{R}$.
I basically got no points when I submitted this approach. In the sample solution our tutor did some "tricky" manipulations where he derived upper and lower bounds to show that $x:=\pm(\alpha n+|y|+1)$, where $\alpha:=\max\{|a_0|,\cdots,|a_{n-1}|\}$ do the job.
However, I can't see where my approach breaks? Maybe someone can explain it to me where I am wrong?
(The question was not to come up with specific values for $x$ it was only about the existence!)
