Why the function is discontinuous at a=1 given #f(x)= 1-x^2# if x < 1 and #f(x)= 1/x# if #x >= 1#?

1 Answer
Oct 11, 2016

The limit as #x# approaches #1# does not exist so it the function is discontinuous at #x=1#.

Explanation:

There are three requirements for a function #f# to be continuous at a #x=a#:

(i) #f(a)# exists
(ii) #lim_(x->a)f(x)# exists
(iii) #f(a)=lim_(x->a)f(x)#

PART 1 We will check if #f(1)# exists. Because #1>=1#, we use #f(x)=1/x#

#f(1)=1/1=1#

#color(blue)(f(a)" exists")#

PART 2 We will check if #lim_(x->a)f(x)# exists. And to do this, we will need to check if the right-hand limit and the left-hand limit are equal.

#lim_(x->a^+)f(x)#
#lim_(x->1^+)1/x=1/1=1#

#lim_(x->a^-)f(x)#
#lim_(x->1^-)(1-x^2)=1-1^2=1-1=0#

Since the right-hand limit and the left-hand limit are not equal, then the limit as #x# approaches #1# does not exist.

The second requirement is not true, therefore the function is discontinuous.