lim x-0 [(3sin4x)/(sin 3x)] = 4 lim x-0 [(sin 4x/4x)x(3x/sin3x)]

How is this true?

Question

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

hartnn · Answer

to get 4x in the denominator, 4x is multiplied in numerator and denominator.
got some idea?

anonymous · Answer

Multiply the whole thing by  (12x/12x).
rearrange.

dumbcow · Answer

you can also use L'Hopitals Rule
-Take derivative of both top and bottom, then reevaluate the limit

anonymous · Answer

But it would be undefined if I took the limit of the bottom

anonymous · Answer

Always with l'Hopital's.   You'd think it was the only thing anyone ever retained from calc. class.

dumbcow · Answer

haha..yeah pretty much

anonymous · Answer

I haven't learned it yet :(

anonymous · Answer

They're learning about sinc(x) here anyway ... that's a few lectures before the class on l'Hopital's.

hartnn · Answer

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got it now?