Answer:
0
Explanation:
![\lim_(n \to 0) (sin^2(x))/(x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gupb4px7cn7rrbxbecui10xcjm8tm0nf0y.png)
When we plug in 0 we will get 0/0
so we apply L' Hopitals rule
We take derivative at the top and bottom
derivative of sin^2(x) is 2sin(x)* cos(x)
2sin(x)cos(x) is sin(2x)
Derivative of x is 1
so limit becomes
![\lim_(n \to 0) (sin(2x))/(1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2kr4iwiss6wu8pfda9esn3s3pvyp20gc30.png)
Plug in 0 for x to find limit
sin(2*0) = 0
So limit value is 0