limit of (sin^2 x)/(x-pi) as x approaches pi

Question

what is the limit?

Answer 1

Question: limit of (sin^2 x)/(x-pi) as x approaches pi?

Let u = x - pi, then as x -> pi, u -> 0.

sin(x) = sin(u+pi) = sin(u).cos(pi) + cos(u).sin(pi) = -sin(u) + 0

sin(x) = -sin(u)

sin^2(x) = sin^2(u)

So, Lim[x->pi] sin^2(x)/(x-pi) = Lim[u->0]sin^2(u)/u

And Lim[u->0]sin^2(u)/u is a standard limit equal to 0.

Answer:  Lim[x->pi] sin^2(x)/(x-pi) = 0

If need be, you can get this (standard) limit by using l'Hopital's rule. Take the limit of the derivative of the numerator over the limit of the derivative of the denominator. You get lim (2sinucosu)/1=2*0*1/1=0

Answer 2

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