L'Hopital's Rule is one way to get the answer. The given limit
#lim_{x->0}(1-cos(2x))/(x^2)# is a "#0/0# indeterminate form" (since both the top and bottom approach zero as #x# approaches zero).
This is a situation where L'Hopital's Rule can be used. Differentiating the top and bottom individually to create a new fraction, the original limit will equal the following limit (if the following limit exists):
#lim_{x->0}(2sin(2x))/(2x)=lim_{x->0}(sin(2x))/x#.
This is, once again, a #0/0# indeterminate form. L'Hopital's Rule can again be used to say it will equal the following limit (if the following limit exists):
#lim_{x->0}(2cos(2x))/1#.
But this limit exists because the function in question is continuous. The limit can be evaluated by substitution: #lim_{x->0}2cos(2x)=2cos(0)=2*1=2#.
Putting these things together implies that #lim_{x->0}(1-cos(2x))/(x^2)=2#.
Another way to get the answer is to use the Taylor (a.k.a. Maclaurin) series for the cosine function expanded about 0. That is, the fact that #cos(z)=1-z^2/(2!)+z^4/(4!)-z^6/(6!)+cdots# for all #z#.
This implies that #(1-cos(2x))/(x^2)=(1-(1-(2x)^2/(2!)+(2x)^4/(4!)-(2x)^6/(6!)+cdots))/(x^2)#
#=(2x^2-2/3 x^4+4/45 x^6-cdots)/(x^2)=2-2/3 x^2+4/45 x^4-cdots# for all #x !=0#.
But the value at #x=0# is irrelevant for the value of the original limit (the original function is undefined at 0). We can say that
#lim_{x->0}(1-cos(2x))/(x^2)=lim_{x->0}(2-2/3 x^2+4/45 x^4-cdots)#
This last function is continuous at #x=0#, so the value of this last function at zero is relevant. We can evaluate the limit by substitution of #x=0# to get
#lim_{x->0}(1-cos(2x))/(x^2)=lim_{x->0}(2-2/3 x^2+4/45 x^4-cdots)=2-0+0-0+0-cdots=2#
