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Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1use lhopitals rule which is \[\frac{\frac{d}{dx}[\sin(2x)]}{\frac{d}{dx}[\sin(3x)]}\]

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1when you do that you will simply get 2/3

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1so your limit will be 2/3

ujuge321
 3 years ago
Best ResponseYou've already chosen the best response.0show me the process. i never learned that in class.

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1what calculus class are you in?

raheen
 3 years ago
Best ResponseYou've already chosen the best response.1using le'hopital rule ==> lim 3co(2x)/3cos(3x) = 2/3

raheen
 3 years ago
Best ResponseYou've already chosen the best response.1using le'hopital rule ==> lim 3cos(2x)/3cos(3x) = 2/3

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3If you don't yet know l'Hopitals rule, I assume you do know already that the limit as x > 0 of (sin x) / x is 1. Now write\[\frac{\sin 2x }{\sin 3x} = \frac{2}{3} \frac{\sin 2x/(2x)}{\sin 3x/(3x)}\] Then you observe that the limits of the ratios of sin over x are all 1 and hence the answer is 2/3.
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