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Prove if $f(x)$ is a polynomials with respective leading terms $ax^{n}$ then $$f(x) \sim ax^{n-m}$$

How do I approach this problem?

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  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$
    – user
    Jan 24 '18 at 21:56
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One way to start is to see if $f(x)/L_f(x)\sim 1$, where $L_f(x)$ is the lead term of $f(x)$.

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  • $\begingroup$ To show something is asymptotic to something else you put one over the other and if the limit as n tends to infinity is 1 then it is asymptotically equal $\endgroup$
    – user511607
    Dec 9 '17 at 15:43
  • $\begingroup$ So would this mean $\lim_{n\to\infty} \frac{f(x)}{ax^{n}}= \lim_{n\to\infty} 1+\frac{a_1 x^{n-1}+a_2 x^{n-2}......+1}{ax^{n}}$ ? $\endgroup$
    – user511607
    Dec 9 '17 at 15:45
  • $\begingroup$ In this problem, it is $x$ that's doing the moving. $\endgroup$ Dec 9 '17 at 15:45
  • $\begingroup$ So $\lim_{x\to\infty} \frac{f(x)}{ax^{n}}= \lim_{n\to\infty} 1+\frac{a_1 x^{n-1}+a_2 x^{n-2}......+1}{ax^{n}}$ how can I show $\lim_{x\to\infty} \frac{a_1 x^{n-1}+a_2 x^{n-2}......+1}{ax^{n}}=0$ $\endgroup$
    – user511607
    Dec 9 '17 at 15:50
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    $\begingroup$ Aah so because $\frac{a_1 x^{n-1}}{ax^{n}}= \frac{a_1}{ax}$ and $\lim_{x\to\infty} \frac{a_1}{ax}=0$ and this can be applied for all the other terms in the series $\endgroup$
    – user511607
    Dec 9 '17 at 15:56
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You have to show that

$$\frac{\frac{f(x)}{g(x)}}{\frac{a}{b}x^{n-m}} \to 1$$

as $x\to \infty$

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