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 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra  
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  
Math Team Joined: Jul 2011 From: Texas Posts: 3,092 Thanks: 1674  Quote:
$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$  
June 12th, 2019, 06:40 AM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out.  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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