anonymous
  • anonymous
find the lim as x approaches zero of 2-2cos(3x)/3x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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raffle_snaffle
  • raffle_snaffle
do you know your limit laws?
anonymous
  • anonymous
some what
raffle_snaffle
  • raffle_snaffle
I guess you could apply L'Hospitals Rule. Did you learn his rule yet?

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More answers

AravindG
  • AravindG
2-2cos(3x)=2(1-cos 3x) Now apply 1-cos t formula
anonymous
  • anonymous
no I have not. I've only learned the basics. like when the numerator exponent is higher the lim does not exist and denominator degree is higher the lim is zero
AravindG
  • AravindG
Try what I said above.
raffle_snaffle
  • raffle_snaffle
Listen to aravinda
AravindG
  • AravindG
*Aravind
anonymous
  • anonymous
ok
anonymous
  • anonymous
I did that what's next?
AravindG
  • AravindG
What did you get?
anonymous
  • anonymous
2(1-COS3X)/3X from this point do I just plug in the zero?
anonymous
  • anonymous
That'll give me 0/0 which its not allowed
anonymous
  • anonymous
please help, I'm very clueless and I can't afford to fell that test tomorrow
Psymon
  • Psymon
Well, in general: \[\lim_{x \rightarrow 0}\frac{ 1-cosx }{ x }=0\]As long as the x angle of cosine and the x int he denominator match, you can say its 0 and NOT undefined. So it looks like you just have 2*0 = 0
Psymon
  • Psymon
@have_sabr
anonymous
  • anonymous
That makes sense Thank you very much! :)
Psymon
  • Psymon
Yep, np :3
Hero
  • Hero
Given: \[\lim_{x \rightarrow 0}\frac{2\cos(3x)}{3x}\] Taking the RIGHT side limit we have: \[\lim_{x \rightarrow 0^{+}}\frac{2\cos(3x)}{3x} = \infty\] Taking the LEFT side limit we have: \[\lim_{x \rightarrow 0^{-}}\frac{2\cos(3x)}{3x}= -\infty\] Therefore, the two sided limit does not exist.

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