Question #368ef
2 Answers
The limit is
Explanation:
Separate this into a product like this:
#= lim_(x->0) (sinx/x)((1-cosx)/tan(pix))#
#= lim_(x->0)(sinx/x) * lim_(x->0)(1-cosx)/tan(pix)#
Use the limit of sin(x) / x rule:
#= 1 * lim_(x->0)(1-cosx)/tan(pix)#
Now use the trig rule
#= lim_(x->0)(2sin^2(x/2))/tan(pix)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since this limit approaches 0, we can see that both
#lim_(x->0)(2sin^2(x/2))/tan(pix) = lim_(x->0)(2sin^2(x))/tan(x)#
#= lim_(x->0)2sinxcosx#
#= 2sin(0)cos(0)#
#= 2(0)(1) = 0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the limit equals
graph{(sinx(1-cosx))/(xtan(pix)) [-1.25, 1.25, -0.625, 0.625]}
Please see below.
Explanation:
Since,
# = (sin^2x cos(pix))/(xsin(pix)(1+cosx))#
# = sin^2x/x^2 * 1/pi * (pix)/sin(pix) * cos(pix)/(1+cosx)#
Taking the limit as
# = 1 * 1/pi * 1 * 1/(1+1) = 1/(2pi)#