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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Please, if you are ok, you can accept the answer and set it as solved. Thanks! – 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 $f(x)$.
kimchi lover
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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 – user511607 Dec 09 '17 at 15:43
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So would this mean $\lim_{n\to\infty} \frac{f(x)}{ax^{n}}= \lim_{n\to\infty} 1+\frac{a_1 x^{n-1}+a_2 x^{n-2}......+1}{ax^{n}}$ ? – user511607 Dec 09 '17 at 15:45
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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 $\lim_{x\to\infty} \frac{a_1 x^{n-1}+a_2 x^{n-2}......+1}{ax^{n}}=0$ – user511607 Dec 09 '17 at 15:50
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1Aah 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 – user511607 Dec 09 '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 $x\to \infty$
user
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