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ujuge321
calculate the following limits: lim sin(2x)/sin(3x) as x approaches to 0.
use lhopitals rule which is \[\frac{\frac{d}{dx}[\sin(2x)]}{\frac{d}{dx}[\sin(3x)]}\]
when you do that you will simply get 2/3
so your limit will be 2/3
show me the process. i never learned that in class.
what calculus class are you in?
using le'hopital rule ==> lim 3co(2x)/3cos(3x) = 2/3
using le'hopital rule ==> lim 3cos(2x)/3cos(3x) = 2/3
If 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.