Question #368ef

2 Answers
John D.
Oct 18, 2017

The limit is #0#

Explanation:

#lim_(x->0) (sinx(1-cosx))/(xtan(pix)#

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 #1 - cosx = 2sin^2(x/2)#

#= lim_(x->0)(2sin^2(x/2))/tan(pix)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Since this limit approaches 0, we can see that both #x/2# and #pix# approach #0# as well. This means that at extremely small values of #x#, we can essentially say that #x ~~ x/2# and #x ~~ pix#, so we can say that:

#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 #0#. This can be confirmed with a graphing calculator:
graph{(sinx(1-cosx))/(xtan(pix)) [-1.25, 1.25, -0.625, 0.625]}

Jim H
Oct 18, 2017

Please see below.

Explanation:

#(sin(1-cosx))/(xtan(pix)) = sin(1-cosx)/((1-cosx)) * (1-cosx)/(xtan(pix))#

Since, #lim_(xrarr0)sin(1-cosx)/(1-cosx) - lim_(urarr0)sinu/u=1#, we will concentrate on the second factor.

#((1-cosx))/(xtan(pix)) * ((1+cosx))/((1+cosx)) = sin^2x/(xtan(pix) ((1+cosx))#

# = (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 #xrarr0# gives

# = 1 * 1/pi * 1 * 1/(1+1) = 1/(2pi)#

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