Question #29a12

1 Answer
Feb 5, 2017

Answer:

#f(x)# is continuous at #x=0#.

Explanation:

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#.