Question

Question #d9c46

Answer 1

`2`

Explanation:

`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

`2.`

Explanation:

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.!