How do you explain why #f(x) = x^2# if #x>=1# and 2x if x<1 is continuous at x = 1?

1 Answer
Nov 30, 2016

Answer:

It is not continuous at #1#, because the limit at #1# does not exist.

Explanation:

#f(x) = {(x^2,"if",x >=1),(2x,"if",x < 1) :}#

#f# is continuous at #1# if and only if

#lim_(xrarr1)f(x) = f(1)#. #" "# (existence of both is implied)

We see from the definition of #f# above that

#f(1) = (1)^2 = 1#.

In finding the limit, we'll need to consider the left and right limits separately because the rule changes at #x=1#.

The limit from the right is

#lim_(xrarr1^+)f(x) = (1)^2 = 1#.

The limit from the left is

#lim_(xrarr1^-)f(x) = 2(1) = 2#.

Because the right and left limits are not the same, the limit does not exist.

Therefore the function is not continuous at #1#.