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Hi Josh. Looking at the fraction, I can see that it goes to zero divided by zero, so my first guess would be to see if I can use l'Hôpital's Rule on it: If, as x approaches 0, lim sin(4x) = 0 [and it does] and lim sin(6x) = 0 [it does] and lim (sin 4x)′ / (sin 6x)′ has a finite value is +/ inifinity [we'll check this], then the lim sin(4x)/sin(6x) = lim (sin 4x)′ / (sin 6x)′. So we need the derivative of sin 4x. Using the chain rule we have (sin 4x)′ = (4x)′ cos(4x) = 4 cos(4x). So lim (sin 4x)′ / (sin 6x)′ = lim [4 cos(4x)] / [6 cos 6x]. As x approaches 0, this converges to 2/3. That's a finite value, so that's the same as the original limit. Hope this helps,  


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