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lim x->0 [(3sin4x)/(sin 3x)] = 4 lim x->0 [(sin 4x/4x)x(3x/sin3x)] How is this true?

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to get 4x in the denominator, 4x is multiplied in numerator and denominator. got some idea?
Multiply the whole thing by (12x/12x). rearrange.
you can also use L'Hopitals Rule -Take derivative of both top and bottom, then reevaluate the limit

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Other answers:

But it would be undefined if I took the limit of the bottom
Always with l'Hopital's. You'd think it was the only thing anyone ever retained from calc. class.
haha..yeah pretty much
I haven't learned it yet :(
They're learning about sinc(x) here anyway ... that's a few lectures before the class on l'Hopital's.
|dw:1347162711646:dw| got it now?
just use sin t / t when limit t->0 = 1 formula.
@hartnn what did you multiply by? It's cut off in the sketch.
Why is it that .... the 4 is moved outside the limit?
multiplied by 4x in num and denom, and it can be seen,iguess
the 4 is constant and constants can be moved outside the limit
Sorry, I meant 4x
so 12x/sin3x
now 12x / sin 3x can again be written as 4 * (3x/sin 3x) and using same limit again its 4*1= 4 ok?
Oh, I get it now! Thank you guys SO MUCH!

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