Question

How do you find `\lim _ { x \rightarrow 0} \frac{ \sin ^ { 2} x }{ 2x }`?

Answer 1

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

Explanation:

Write the function so that we can use the well-known limit of `sinx/x`:

`lim_(x->0) sin^2x/(2x) = lim_(x->0) [1/2* sinx *(sinx/x)] = 1/2*0*1=0`