We have to find lim x --> 0 [(1- cosx)/x^2]

cos x = 1 - 2*(sin x/2)^2

=> 1 - cos x = 2*sin(x/2)^2

lim x --> 0 [(1- cosx)/x^2]

=> lim x -->0 [2*(sin x/2)^2/x^2]

On substituting x = 0 we get the indeterminate form 0/0,

So we can use l'Hopital's Rule and substitute the numerator and denominator by their derivatives

=> lim x-->0[ 2*cos(x/2)*sin(x/2)/ 2x]

=> lim x-->0 [ sin x / 2x

Again x = 0 gives us 0/0, so repeat the procedure again

=> lim [cos x / 2]

substitute x = 0, we get 1/2

**The required limit is 1/2**

We'll substitute the numerator:

1 - cos x = 2 [sin(x/2)]^2

Lim (1-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2

Lim {2 [sin(x/2)]^2}/x^2 = Lim [sin(x/2)]^2/(x^2/2)

Lim [sin(x/2)]^2/(x^2/2) = lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)]

Since the elementary limit is:

lim [sin f(x)]/f(x) = 1, if f(x) -> 0, we'll get:

**Lim (1-cosx)/x^2 = 1/2**