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clara1223

  • one year ago

find the limit as x approaches 0 of (sin^2)(4x)/3x

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  1. anonymous
    • one year ago
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    Do you know L'Hopital's Rule?

  2. clara1223
    • one year ago
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    @PlasmaFuzer no

  3. jim_thompson5910
    • one year ago
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    Hint: \[\Large \lim_{x \to 0}\left(\frac{\sin(x)}{x}\right) = 1\]

  4. clara1223
    • one year ago
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    Oh yes i know that

  5. clara1223
    • one year ago
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    I'm just not sure how to simplify it to make it look like that

  6. anonymous
    • one year ago
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    Ahh yes squeeze theorem good one @jim_thompson5910

  7. freckles
    • one year ago
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    \[\sin^2(u)=\sin(u) \cdot \sin(u)\]

  8. anonymous
    • one year ago
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    I've gotten so used to L'Hopital's rule I'm lazy

  9. jim_thompson5910
    • one year ago
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    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

  10. anonymous
    • one year ago
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    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

  11. clara1223
    • one year ago
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    @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?

  12. jim_thompson5910
    • one year ago
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    there's no need because you can simply plug in x = 0 for that first fraction

  13. clara1223
    • one year ago
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    so then would the answer be \[\frac{ 4 }{ 3 }\times0\times1\]

  14. jim_thompson5910
    • one year ago
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    correct

  15. clara1223
    • one year ago
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    Great! so the answer is 0. thanks!

  16. jim_thompson5910
    • one year ago
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    you're welcome

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