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I was give this to prove: f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ has a limit of $\lim_{x \rightarrow \infty} $a>0$

This is the proof I have provided. Please bear with me, I have never written a proper proof so I am going with my gut on writing this proof. I am sorry if this makes some of your stomachs turn.

Proof

Since $a_n > 0$ therefore, we can assume that $x^n$ is even or odd and furthermore, can state the graph of the function will either be even or odd.

If we state that the graph of the function is either even or odd then we can say that they must look like this or some variation of this for higher $n$ value and the cannot be reflect on the $x$ axis because $a \gt 0$.

simple x square graph or x cube graph

Since both of these graphs indicate that as $x\rightarrow\infty$ so does $f(x)$ Therefore, the limit of the function in either even or odd case must be $\infty$. Therefore the $\lim_{x\rightarrow\infty} f(x) = \infty$

Would this be a good enough or a correct proof?

EDIT

I followed the comments and this is what I have done.

$$\lim_{x\rightarrow\infty}a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 $$ $$ \lim_{x\rightarrow\infty} x^n (a_n + \frac{a_{n-1}}{x^n} + \cdot + \frac{a_1}{x^n} + \frac{a_0}{x^n}) $$ $$ \lim_{x\rightarrow\infty}a_nx^n = \infty$$ $$ \lim_{x\rightarrow\infty}f(x) = \infty$$

Thanks!

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No, you're essentially assuming the result. Hint: factor $x^n$ out from the polynomial. Then you have $f(x)= x^n(a_n+\bigl(\text{ stuff that tends to zero as } x\rightarrow\infty)\bigr)$. (Of course, you'll need to explain why that "stuff" actually tends to zero, and why $f(x)$ tends to $\infty$.) –  David Mitra Feb 8 '13 at 0:46
    
@gekkostate I don't think this is considered as a proof, because one can argue that the graphs above are the graphs of $x^n$ but not the graphs of $f(x)$ so one can say you "proved" that $x^n\rightarrow \infty$ but you did not prove any thing about $f(x)$. –  i.a.m Feb 8 '13 at 0:48
    
Re your edit: The powers of $x$ in the denominators are off; the third equation in the display would be better written as $\lim\limits_{x\rightarrow\infty} (a_nx^n)=\infty$. Note that it is here where you use the fact that $a_n>0$. –  David Mitra Feb 8 '13 at 1:06
    
@gekkostate you have a little mistake in what you wrote based on the comment of David, when taking $x^n$ as a common factor you will not get $\frac{a_{n-1}}{x^n}$ and when you concluded the result it seems that $a_n$ did not play a role! –  i.a.m Feb 8 '13 at 1:09

2 Answers 2

up vote 2 down vote accepted

If you know that $$\lim_{x\rightarrow \infty}\frac{a}{x^n}=0$$ for all $n\ge 1$ and for all $a\in \mathbb R$ then you will have$$\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}x^n(a_n+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n})$$ which implies $$\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}x^n(\lim_{x\rightarrow \infty}(a_n+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}))=a_n\lim_{x\rightarrow \infty}x^n=\infty$$ since $a_n> 0.$

Note: you can also see that if $a_n<0$ then you will get $$\lim_{x\rightarrow \infty}f(x)=-\infty.$$

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Sorry, can you please explain your third line a little more. I'm getting confused with the $a_n$ portion and it's significance. –  Jeel Shah Feb 8 '13 at 1:16
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@gekkostate we know that $lim_{x\rightarrow \infty}x^n=\infty$ when you multiply $x^n$ by a positive number $a_n$ then $lim_{x\rightarrow \infty}a_{n}x^n$ does not change, but for example if you multiplied $x^n$ by $-1$ then $lim_{x\rightarrow \infty}-x^n=-\infty$. –  i.a.m Feb 8 '13 at 1:19
    
ahh I get it. Thank you –  Jeel Shah Feb 8 '13 at 1:21
    
you are welcome. –  i.a.m Feb 8 '13 at 1:21

sorry but a world of 'no!!!!' and a couple of 'never'.

You can't rely of schematic graphs for rigorous proofs. More importantly, your claim that $n$ in $x^n$ is either even or odd is a vacuous statement. Much more importantly, your claim that therefore the function at hand will either be even or odd is completely wrong. Consider $x^2+x-76$.

Hint: find the definition of limit and work with it.

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