clara1223
  • clara1223
find the limit as x approaches 0 of (sin^2)(4x)/3x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Do you know L'Hopital's Rule?
clara1223
  • clara1223
@PlasmaFuzer no
jim_thompson5910
  • jim_thompson5910
Hint: \[\Large \lim_{x \to 0}\left(\frac{\sin(x)}{x}\right) = 1\]

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More answers

clara1223
  • clara1223
Oh yes i know that
clara1223
  • clara1223
I'm just not sure how to simplify it to make it look like that
anonymous
  • anonymous
Ahh yes squeeze theorem good one @jim_thompson5910
freckles
  • freckles
\[\sin^2(u)=\sin(u) \cdot \sin(u)\]
anonymous
  • anonymous
I've gotten so used to L'Hopital's rule I'm lazy
jim_thompson5910
  • jim_thompson5910
I suggest first breaking up the square as freckles wrote \[\Large \frac{\sin^2(4x)}{3x}=\frac{\sin(4x)}{3}*\frac{\sin(4x)}{x}\] then try to make the denominator x turn into 4x
anonymous
  • anonymous
Btw once you have completed the problem using jim's method I would suggest you check this out: https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule as you will eventually use it. It makes problems like this just a simple matter of taking derivatives. Though if you haven't seen it in class I would recommend against using it in the solution for this problem in particular, although if you fell able you should try it out and compare the effort it takes using both methods :D
clara1223
  • clara1223
@jim_thompson5910 so I am able to get to \[\frac{ 4 }{ 3 }\lim_{x \rightarrow 0}\frac{ \sin x }{ 1 } \times \frac{ \sin x }{ x }\] but how do I get an x into the first denominator?
jim_thompson5910
  • jim_thompson5910
there's no need because you can simply plug in x = 0 for that first fraction
clara1223
  • clara1223
so then would the answer be \[\frac{ 4 }{ 3 }\times0\times1\]
jim_thompson5910
  • jim_thompson5910
correct
clara1223
  • clara1223
Great! so the answer is 0. thanks!
jim_thompson5910
  • jim_thompson5910
you're welcome

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