rainbow22
lim x>0 [(3sin4x)/(sin 3x)] = 4 lim x>0 [(sin 4x/4x)x(3x/sin3x)]
How is this true?



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

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

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

rainbow22
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But it would be undefined if I took the limit of the bottom

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

dumbcow
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haha..yeah pretty much

rainbow22
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I haven't learned it yet :(

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

hartnn
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dw:1347162711646:dw
got it now?

hartnn
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just use sin t / t when limit t>0 = 1 formula.

Algebraic!
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@hartnn
what did you multiply by? It's cut off in the sketch.

rainbow22
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Why is it that .... the 4 is moved outside the limit?

hartnn
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multiplied by 4x in num and denom, and it can be seen,iguess

hartnn
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the 4 is constant and constants can be moved outside the limit

rainbow22
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dw:1347163083569:dw

rainbow22
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Sorry, I meant 4x

rainbow22
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so 12x/sin3x

hartnn
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now 12x / sin 3x can again be written as 4 * (3x/sin 3x)
and using same limit again its 4*1= 4
ok?

Algebraic!
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dw:1347163221313:dw

rainbow22
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Oh, I get it now!
Thank you guys SO MUCH!