lim[x→π]sin²x / (x-π) ··· Eq.1
Define the numerator as f(x) = sin²x. So, we have: f(π) = sin²π = 0, f'(x) = 2sinx·cosx = sin2x and f'(π) = sin2π = 0 Therefore, Eq.1 can be restated as follows: lim[x→π]sin²x / (x-π) = lim[x→π]f(x) / (x-π) = lim[x→π]{f(x)-0} / (x-π) ··· Eq.2
Since f'(π) exists, we can use the definition of the derivative of f(x), such as, lim[x→a]{f(x)-f(a)} / (x-a) = f'(a)
So that,Eq.2 can be restated as follows: lim[x→π]{f(x)-0} / (x-π) = lim[x→π]{f(x)-f(π)} / (x-π) = f'(π) = sin2π = 0
Therefore, we have: lim[x→π]sin²x / (x-π) = 0