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 June 4th, 2019, 07:29 AM #1 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. asymptotics If $$\lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n)$$ ?....
 June 4th, 2019, 08:16 AM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 817 Thanks: 113 Math Focus: Elementary Math If you set $\displaystyle a/b$ inside the limit then it is done .
 June 4th, 2019, 08:38 AM #3 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. no but then why define $\sim$ the way it is
June 4th, 2019, 09:09 AM   #4
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Quote:
 Originally Posted by Joppy If $$\lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n)$$ ?....
$a=0$, $f(x)=x$, $g(x)=x^2$

 June 7th, 2019, 12:02 AM #5 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. No I get it's a stupid question but I meant $a \ne 0$ also. Basically I have a function with known limit which is not unity and I want to infer how it grows without having to juggle terms in the limit
 June 12th, 2019, 01:47 AM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. The point is that the limit either a) converges to a constant greater than 0, diverges to $\pm \infty$ or converges to 0. If a) then $f \sim g$, b) $f$ grows faster, c) $g$ grows faster.
June 12th, 2019, 06:21 AM   #7
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Quote:
 Originally Posted by Joppy If $$\lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n)$$ ?....
$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$

$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$

June 12th, 2019, 06:40 AM   #8
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 Originally Posted by skeeter $\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$ $\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$
Yes.. just in the asymptotic sense, the constants $a,b$ don't factor into the expression $f \sim g$ because they're negligible in the limit.

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