`lim_(x to 0) (sin (pi cos^2 x))/x^2` बराबर है :

Options

A. `2-pi`
B. `pi`
C. `pi/2`
D. 1

Answer

B

Explanation

`sin^2 x + cos^2 x = 1`

` cos^2 x = 1 - sin^2 x`

`lim_(x-> 0)(sin(pi- pi sin^ 2 x) )/x^ 2`

चुकी

`sin(180 - theta) = sin theta`

`lim_ (x-> 0)(sin(pi * sin^2 x))/(x ^ 2) `

`lim_ (x-> 0) (sinf(x) /f(x) =1)`

`f(0)=0`

`lim_( x-> 0) (sin(pi sin^ 2 x))/( pi sin^ 2 x )* (pi sin^ 2 x)/( x^ 2)`

`1 xx pi xx 1 = pi`