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  • 5 years ago

Regarding convergence and divergence I dont understand from 1. I mean I know 2sin^2x=2sinxcosx but how do we diff them and how come from 1 turn into 2 and 3, the red number 1 is my 2nd question http://imageshack.us/photo/my-images/33/calculusquestion.jpg/

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  1. anonymous
    • 5 years ago
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    Since sin(1/k)^2 is just sin(1/k)*sin(1/k), you can use the product rule to differentiate. So the derivative is just (-1/k^2)*cos(1/k)*sin(1/k)+(-1/k^2)*sin(1/k)*cos(1/k)=2(-1/k^2)*sin(1/k)*cos(1/k). I think you know how to differentiate 1/k^2, but if you don't rewrite 1/k^2 as k^-2 so that you can use the power rule which will give you -2*k^-3=-2/k^3. Now you can simplify [2(-1/k^2)sin(1/k)cos(1/k)]/(-2/k^3) by realizing that (-1/k^2)/(-2/k^3)=2/k. Then you have to recall the double angle formulas. sin(2x)=2sin(x)cos(x). Setting x=1/k you get sin(2/k)=2sin(1/k)cos(1/k). That brings you to step 3. Lim as x approaches 0 of sin(x)/x is equal to one is just something my calc teacher told me to memorize. He didn't teach us how to prove it but I think it has something with the squeeze theorem. I attached problem 2

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