We say that a function f(x) is continuous at x_0 if
- f(x_0) exists
- lim_(x->x_0^+)f(x) = f(x_0)
- lim_(x->x_0^-)f(x) = f(x_0).
For the given f(x) and x_0 = 0, the first condition is satisfied as f(0) = 1/2. To evaluate the limits, observe that f(x) is identical to the function (1-cos(x))/x^2 at all points other than 0, and thus we may work on that function when calculating the limit as f(x) approaches 0.
To evaluate lim_(x->0^+)(1-cos(x))/x^2, we note that 1-cos(x) and x^2 are differentiable on an open interval with the endpoint 0, for example, (0, 1). Additionally, d/dxx^2!=0 and direct substitution of x=0 leads to an indeterminate form 0/0. Thus we may apply L'Hopital's rule.
lim_(x->0^+)(1-cos(x))/x^2 = (d/dx(1-cos(x)))/(d/dxx^2)
=lim_(x->0^+)sin(x)/(2x)
=1/2lim_(x->0^+)sin(x)/x
=1/2(1)
The above follows from the well known limit lim_(x->0)sin(x)/x = 1. It may also be verified via a second application of L'Hopital's rule.
=1/2
=f(0).
Verification of the limit lim_(x->0^-)f(x) = 1/2 follows in much the same manner.
As all three conditions are satisfied, we can say that f(x) is continuous at x=0.
