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Mathematics 18 Online
OpenStudy (anonymous):

evaluate limit -> 0 sin^2(pix) / 3x^2? the pi throws me off

OpenStudy (turingtest):

as x goes to zero?\[ \lim_{x \rightarrow 0}[\sin^2 (\pi x)/3x^2]\] like so?

OpenStudy (anonymous):

yess

OpenStudy (anonymous):

take derivative from nominator and denominator 2sin(pix)*pi*cos(pix) / 6x again : (pi/3) * ((picos^2(pix) - pisin^2(pix)/1) (pi^2 / 3)

OpenStudy (turingtest):

use L'Hospital's Rule I think derivative of top and bottom: \[2 \pi \sin (\pi x)\cos (\pi x)\] (top) 6x on the bottom

OpenStudy (turingtest):

Coolsector is right. I got the same

OpenStudy (anonymous):

you can as well : based on the limit lim x->0 sin(kx)/kx = 1 multiply by pi^2/pi^2 to get : pi^2sin^2(pix) / 3pi^2x^2 now : pi^2/3 * lim(sin(pix)/pix)) * (lim(sin(pix)/pix) = pi^2/3

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