# If $\displaystyle\lim_{n\to\infty} x_n = +\infty$ then $\displaystyle\lim_{n\to\infty} P(x_n) = \lim_{x\to\infty} P(x)$.

Is there any problem with expressing that, if $\displaystyle\lim_{n\to\infty} x_n = +\infty$ for some sequence $\{x_n\}_{n=1}^\infty$ then $$\lim_{n\to\infty} P(x_n) = \lim_{x\to\infty} P(x)?,$$ where $P$ is a polynomial of the form $P(x)=a_0+...+a_n x^n$. How should I formalize these reasoning? Any hint for proving that $\lim_{n\to\infty} P(x_n) = \pm \infty$ if there exists $a_r\neq 0$ with $r\in\{1,...,n\}$? Thanks in advance and sorry for editing the question.

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Well, that changed the question considerably. Strictly speaking though, $\lim_{x \to \infty} f(x)$ doesn't exist if $f$ is a polynomial (unless $f$ is constant). I assume you allow improper limits here? – mrf Feb 26 '13 at 20:52
@mrf You're right, that's okay. I'm only trying to prove that $\lim_{n\to\infty} f(x_n) =\pm\infty$. – V. Galerkin Feb 26 '13 at 20:56

If you already know that $\lim_{x\to\infty} f(x)$ exists (possibly as $\pm\infty$), it's true that $\lim_{x\to\infty} f(x_n) = \lim_{x\to\infty} f(x)$ for every sequence $x_n$ tending to $\infty$. (Whether $f$ is a polynomial or not.)
Just look at the definition of limit to see why this is true: By assumption $|f(x)-L| < \varepsilon$ for $x$ large enough. Hence $|f(x_n)-L| < \varepsilon$ for $n$ large enough (since $x_n \to \infty$ as $n\to\infty$.) Adapt the reasoning when $L = \pm\infty$.