What is the limit of #(sin^2x)/(3x^2)# as x approaches #0#?

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Answer 1 · 2016-04-05T20:49:57

#lim_(x->0) sin^2(x)/(3x^2) = 1/3#

Start with your favourite proof that #lim_(x->0) (sin(x))/x = 1#

That might start with a geometric illustration that for small #x > 0#

#sin(x) <= x <= tan(x)#

Then divide through by #sin(x)# to get:

#1 <= x / sin(x) <= 1 / cos(x)#

Take reciprocals and reverse the inequality (since #1/x# is monotonically decreasing for #x > 0#) to get:

#cos(x) <= sin(x)/x <= 1#

Then #lim_(x->0+) cos(x) = 1#

So #lim_(x->0+) (sin(x)/x) = 1#

Also #sin(-x) = -sin(x)#, so #sin(-x)/(-x) = sin(x)/x#

Hence #lim_(x->0-) (sin(x)/x) = 1# too.

#color(white)()#
Whatever method we use to find #lim_(x->0) sin(x)/x = 1#, it is not too difficult to deduce:

#lim_(x->0) sin^2(x)/(3x^2) = (lim_(x->0) sin(x)/x) * (lim_(x->0) sin(x)/x) * 1/3 = 1/3#