If $P(x)$ is a polynomial then $\lim_{x \to \infty }P(x) = 0 \iff P(x)=0$, How to show that if $P(x)$ is a polynomial then $\lim_{x \to \infty }P(x) = 0 \iff P(x)=0$,
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 A: We call the degree of a polynomial $n$.
For $n \ge 1$ $$P(x) = c x^n + O(x^{n-1})$$ (for some nonzero $c$) so limit as $x \to \infty$ is $\pm \infty$ depending on the sign of $c$.
Therefore if $\lim x \to \infty$ gives $0$ then $n = 0$, i.e. $P(x) = 0$.
A: if $p(x)=0$ it is clear, now for the converse we have$$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$$ then $$p(x)=x^n(a_n+\frac{a_n-1}{x}+...+\frac{a_0}{x^n})$$ now take the limit as $x\to \infty$ you will get that $$\lim_{x\to \infty}a_nx^n=0$$ since $\lim_{x\to\infty}p(x)=0$ but this is true iff $a_n=0$ ,now you can do the same to get $a_k=0$ where $k\in \{0,...,n-1\}$. Thus, we must have p(x)=0.
A: Call a non-zero polynomial $P(x)$ weird if $\lim_{x\to \infty}P(x)=0$. Suppose there are no weird polynomials of degree $k$. We show there is no weird polynomial of degree $k+1$. 
This is easy: if $P(x)$ is weird of degree $k+1$, then $P(x+1)-P(x)$ is weird of degree $k$. 
A: That $\lim 0=0$ when $x\to \infty$ is immediate. You can prove the converse pretty easily. If $P\neq 0$ then $\lim P\neq 0$ when $x\to\infty$.
A: Suppose $P(x) \to 0$ as $x \to \infty$. Let $P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} +\cdots + a_0$. For $x \neq 0$, we can define
$$Q(x) = \frac{P(x)}{x^n} = a_{n} + a_{n-1}x^{-1} + \cdots + a_{0}x^{-n}$$
As $x \to \infty$, $x^{-1},x^{-2}, \cdots, x^{-n}$ all go to zero, so $Q(x) \to a_n$. If $P(x) \to 0$, then we also know that $Q(x) \to 0$. So $a_n = 0$.
A: It is clear that if $P(x)=0$ then the limit as $x$ goes to infinity is $0$.
To prove the converse by contradiction one could do the following. First assume $P(x)$ has positive degree. Therefore the limit as $x$ goes to infinity cannot be $0$ (as $P(x)$ is not bounded) so $P(x)$ must be constant. Finally, it is clear that the only constant polynomial satisfying the required condition is the $0$ polynomial.
