anonymous
  • anonymous
lim x->0 [(3sin4x)/(sin 3x)] = 4 lim x->0 [(sin 4x/4x)x(3x/sin3x)] How is this true?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
to get 4x in the denominator, 4x is multiplied in numerator and denominator. got some idea?
anonymous
  • anonymous
Multiply the whole thing by (12x/12x). rearrange.
dumbcow
  • dumbcow
you can also use L'Hopitals Rule -Take derivative of both top and bottom, then reevaluate the limit

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anonymous
  • anonymous
But it would be undefined if I took the limit of the bottom
anonymous
  • anonymous
Always with l'Hopital's. You'd think it was the only thing anyone ever retained from calc. class.
dumbcow
  • dumbcow
haha..yeah pretty much
anonymous
  • anonymous
I haven't learned it yet :(
anonymous
  • anonymous
They're learning about sinc(x) here anyway ... that's a few lectures before the class on l'Hopital's.
hartnn
  • hartnn
|dw:1347162711646:dw| got it now?
hartnn
  • hartnn
just use sin t / t when limit t->0 = 1 formula.
anonymous
  • anonymous
@hartnn what did you multiply by? It's cut off in the sketch.
anonymous
  • anonymous
Why is it that .... the 4 is moved outside the limit?
hartnn
  • hartnn
multiplied by 4x in num and denom, and it can be seen,iguess
hartnn
  • hartnn
the 4 is constant and constants can be moved outside the limit
anonymous
  • anonymous
|dw:1347163083569:dw|
anonymous
  • anonymous
Sorry, I meant 4x
anonymous
  • anonymous
so 12x/sin3x
hartnn
  • hartnn
now 12x / sin 3x can again be written as 4 * (3x/sin 3x) and using same limit again its 4*1= 4 ok?
anonymous
  • anonymous
|dw:1347163221313:dw|
anonymous
  • anonymous
Oh, I get it now! Thank you guys SO MUCH!

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