Simple trig limit with cos

  • Thread starter L4N0
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  • #1
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Homework Statement


Evaluate the limits that exist:

lim x->0 of [tex]\frac{1-cos4x}{9x^{2}}[/tex]

Homework Equations


lim x->0 [tex]\frac{1-cosax}{ax}[/tex] = 0


The Attempt at a Solution


so far I've got this
[tex]\frac{4x}{9x^{2}}\frac{1-cos4x}{4x}[/tex]
The second part has a limit of 0 but I don't know what to do about the first fraction. To me it looks undefined but according to some java applet I found on the internet the limit should be 8/9. What's the right answer?
 

Answers and Replies

  • #2
Dick
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The limit has the form 0/0. That suggests you use L'Hopital's rule. Alternatively, if you know the power series expansion of cos(x)=1-x^2/2!+x^4/3!-..., you could substitute x->4x and use that.
 
Last edited:
  • #3
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Thank you. So far we are in Ch 2. so we haven't done L'Hopital's rule or power series. I guess saying it is undefined is enough for now.
 
  • #4
Dick
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The trouble with that is that is IS defined. If you can't use those, then you have to use tricks. Try a double angle formula like cos(2x)=1-2*sin(x)^2. Use that to express cos(4x). Then use lim x->0 sin(ax)/(ax)=1. I'm guessing you do know that from the relevant equation you posted.
 
  • #5
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oops, I got confused a bit...
Anyways, I used your suggestion and solved it.
Thanks again.
 
  • #6
dynamicsolo
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oops, I got confused a bit...
Anyways, I used your suggestion and solved it.
Thanks again.

Very sneaky little problem: the limit law it looks the most obvious to use is the wrong one!. Keep these kinds of manipulations in mind on exam problems...
 

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