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Hello I'm trying to prove the following fact

Let $f(x) = x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ with $n \in \mathbb{N}$ and $a_{n-1},\ldots,a_0 \in \mathbb{R}$. If $n$ is even, then we have $\displaystyle\lim_{x\to+\infty}f(x)=+\infty$ and $\displaystyle\lim_{x\to -\infty}f(x)=+\infty$.

This seems quite intuitive since the highest degree of the polynomial and any number squared is even. However this is obviously not a proof. So my second thought was that there could be a way to prove that $f(x)$ is not injective, meaning that some values could have more than one corresponding $x$ value. However, this also does not quite hold (I think) for any polynomial. However this is where I'm stuck. How can I properly prove it ?

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1 Answer 1

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Hint: Use the fact that $$f(x)=x^{n}\left[1+\frac {a_{n-1}} x+\frac {a_{n-2}} {x^{2}}+\cdots+\frac {a_0} {x^{n}}\right].$$

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  • $\begingroup$ Oh so i can just say that the limit of each of these terms is 0 (except for the 1 of course) and then go on from there saying $x^n$ with $n$ even. Should I try proving that using the property that $f(x) = f(-x)$ for all $x$? Although I'm not quite sure how that would be done $\endgroup$
    – 23408924
    Nov 9, 2020 at 9:31
  • $\begingroup$ Exactly like that @23408924. $\endgroup$
    – Souza
    Nov 9, 2020 at 9:35
  • $\begingroup$ If $g(x) \to \infty$ and $h(x) \to 1$ then $g(x)h(x) \to \infty$. $\endgroup$ Nov 9, 2020 at 9:37
  • $\begingroup$ Okay thank you I will try to prove it this way $\endgroup$
    – 23408924
    Nov 9, 2020 at 9:43

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