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OpenStudy (anonymous):
lim x->0 sin3x sin5x/x^2
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OpenStudy (owlfred):
\[\sin (3x)\sin(5x) = \frac{\cos(2x) - \cos(8x)}{2}\]
OpenStudy (owlfred):
Or just use L'Hospital's.
OpenStudy (anonymous):
ah oh okay got it thank you
OpenStudy (owlfred):
:)
OpenStudy (xapproachesinfinity):
you can also separate that x^2=x*x then multiply and divide by 3 as well as 5 then you have the usual limit sinx/x at zero which is 1 and the numbers that you multiplied by are what remains
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