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        riddles >> putnam exam (pure math) >> Probably easy limit question
        
(Message started by: Lizzy on Mar 20^th, 2006, 7:00pm)

Title: Probably easy limit question
Post by Lizzy on Mar 20^th, 2006, 7:00pm

Hey guys, this will probably be really easy for u...but i don't know how to do it :(

Find the limit of

cot(2x)sin(6x) as x approaches 0

any help on how to do it will be greatly appreciated

Lizzy

xoxo

Title: Re: Probably easy limit question
Post by Barukh on Mar 20^th, 2006, 11:41pm

Making a substitution z = 2x, you simplify the limit to cot(z)sin(3z), so actually you need to find the limit of sin(3z)/sin(z). Use the trigonometric identity for a triple angle to obtain the answer 3.

Another way is to look at the power series for cot(x) and sin(x).

Title: Re: Probably easy limit question
Post by Icarus on Mar 21^st, 2006, 3:45pm

Another approach: recall the identities lim_z->0 (sin z)/z = 1 and lim_z->0 (tan z)/z = 1.

The second can be rewritten as lim_z->0 z cot z = 1. So,

lim cot 2x sin 6x = lim 3(2x cot 2x)((sin 6x)/6x)
= 3 (lim 2x cot 2x)(lim (sin 6x)/6x) = 3 * 1 * 1 = 3.

Title: Re: Probably easy limit question
Post by Lizzy on Mar 22^nd, 2006, 2:29am

yay! i understand icarus' method :D

hmm, does anyone know any good online resources to learn this stuff properly?

Title: Re: Probably easy limit question
Post by Icarus on Mar 22^nd, 2006, 6:19pm

Sorry, but I don't. I learned it before the internet was easily accessible.