anonymous
  • anonymous
lim x→0 cos 9x − cos^2 *9x / 9x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[lim~x\rightarrow \infty\{ \cos(9x) -\frac{\cos^2(9x)}{9x}~\text{?}\]
anonymous
  • anonymous
on the homework its setup like lim x -> 0 ........ cos(9x) - cos^2 9x / 9x
campbell_st
  • campbell_st
break it into smaller parts... 1. \[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} \cos^2(9x)/9x\]

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anonymous
  • anonymous
okay, well you can separate this into two distinct limits, one is very easy to deal with because we know what it turns out to be: like this \[lim~ x \rightarrow 0~~cos(9x) - lim~x \rightarrow 0~~\frac{cos^2(9x)}{9x)}\] the other one resembles the sin function you asked about a few questions ago, the same methods should work
anonymous
  • anonymous
atleast I think it was you that asked about it
anonymous
  • anonymous
so should it be 1/9 0r 1/81
campbell_st
  • campbell_st
the cos^2(9x) may use l'hopitals rule...
anonymous
  • anonymous
im thinking the 9's match like cos(9x)/ 9x = 1 and then cos 9x *cos 9x / 9x = 1 am i on the right track?
anonymous
  • anonymous
campbell, no you can't you don't have a case of 0/0,0/infinity,infinity/0 or infinity/infinity, you have 1/0
campbell_st
  • campbell_st
well lim of cos(9x) = 1 as x ==> 0
anonymous
  • anonymous
is it 3?
anonymous
  • anonymous
if you try plotting this one you can see pretty clearly that as x -> 0 the function approaches infinity. there's another pretty useful property of limits that says the limit of a product of two functions is the product of the limits. so what you'd have in this case is \[1-\{lim~x\rightarrow 0~(\cos^2(x))\}*\{lim~x\rightarrow0(\frac{1}{9x})\}\] the limit of one of these is one, while the other is infinity, and 1 times infinity is infinity.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=limit+as+x+-%3E0+of+cos^2%28x%29%2Fx
anonymous
  • anonymous
I dont think its infinity, i put it into the thing at it said an x, but i think its 3
anonymous
  • anonymous
but its 9(x)
campbell_st
  • campbell_st
use the idea that that the sum of the limits \[\cos ^{2}(9x) = 1 - \sin^2(9x)\] then is \[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} (1 - \sin^2(9x))/9x\]
campbell_st
  • campbell_st
\[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} 1/9x + \lim_{x \rightarrow \infty} \sin^2(9x)/9x\]

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