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Answer 1 · 2017-01-28T17:23:51

#2#

#sin^2(2x)/(2x^2) = 2sin^2(2x)/(2x)^2=2(sin(2x)/(2x))^2# so

#lim_(x->0)sin^2(2x)/(2x^2) =2(lim_(x->0)sin(2x)/(2x))^2=2cdot1^2=2#

Answer 2 · 2017-01-28T17:32:01

#2.#

We will use the following Standard Form of Limit :-

#lim_(thetato0)sintheta/theta=1#.

Put #2x=y :." as "x to 0, y to 0." Also, "x=y/2#.

#"Now, the Reqd. Lim.="lim_(y to 0) sin^2y/(2(y^2/4)#

#=lim_(y to 0) (2sin^2y)/y^2=2[lim_(y to 0) (siny/y)^2]=2(1)^2=2.#

Enjoy Maths.!