Re: limits

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#64

From: Robert Morewood
Date: Mon, 07 Dec 1998 00:00

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pez (hot@mail.com) wrote:
: how do you do this?
: 
: lim    [tan(3x)]/2x
: x->0
: 
: I thought of using
: 
: lim    sin(3x)/[cos(3x)2x]
: x->0
: 
: but then what?

You should know:
 lim    sin(x)/x
 x->0

So you also know:
 lim    sin(3x)/[3x]
 x->0

You only have TWO x in your denominator, so multiply the top and bottom
by THREE.  Then you can factor out  sin(3x)/[3x].  What's left is easy.

Check your answer by calculating Tan(3x)/(2x) for x=0.1,0.01,0.001,...
(Just be careful, some calculators start to give wrong answers when
x gets too close to zero!)

  |\/|                          ||  Burnaby South Secondary School
  |  |orewood@planeteer.com     ||    Beautiful British Columbia
Mathematics & Computer Science  ||             (Canada)

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From: Robert@van1p8.planeteer.com (Robert Morewood)
Subject: Re: limits
Date: 1998/12/07
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