xilinix
lim -> 0 sin(2x) /(x*cos(x) )
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myko
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sin (2x) = 2sin(x)cos(x)
so sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x
which as x->0 goes to 1
xilinix
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no, answer is 2
myko
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ups sry, forgot about 2 multuplying:
lim 2sinx/x=2*1=2
x->0
xilinix
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don't undestand you
myko
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sin (2x) can be written like:
sin (2x) = 2sin(x)cos(x)
substituting this expretion in the original one:
sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x
taking limits, and noticing that lim as x->0 of sinx/x=1:
lim 2sinx/x=2*1=2
x->0
xilinix
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|dw:1352667129469:dw|
myko
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no
xilinix
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why not? l'Hôpital rule
ccappiello
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yes, using l'hospital's rule works
Varuns
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Don't break your head ! You know that sin(x)=x as x tends to zero. So sin(2x) = 2x. In the denominator, keep x as it is and cos(x)=1. Hence the answer
xilinix
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thanks, but the image is ok?
Varuns
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|dw:1352669528245:dw|
lopus
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claro que es posible es solamente derivar arriba y abajo como lo muestras en la imagen y ya