Question #2e37e

1 Answer
Nov 12, 2017

This function is not continuous at #x=0# or at #x=pi#. The left- and right-hand limits are not equal.

Explanation:

The left-hand limit as #x# approaches zero from the left uses the first formula: #lim_{x->0-}f(x)=lim_{x->0-}e^{x}=e^{0}=1#.

The right-hand limit as #x# approaches zero from the right uses the second formula: #lim_{x->0+}f(x)=lim_{x->0+}sin(x)=sin(0)=0#.

Since #1!=0#, the two-sided limit #lim_{x->0}f(x)# does not exist, so #f# is not continuous at #x=0# (there is a "jump discontinuity" at #x=0#).

The left-hand limit as #x# approaches #pi# from the left uses the second formula: #lim_{x->pi-}f(x)=lim_{x->pi-}sin(x)=sin(pi)=0#.

The right-hand limit as #x# approaches #pi# from the right uses the third formula: #lim_{x->pi+}f(x)=lim_{x->pi+}(x-pi-1)=pi-pi-1=-1#.

Since #0!=-1#, the two-sided limit #lim_{x->pi}f(x)# does not exist, so #f# is also not continuous at #x=pi# (there is also a "jump discontinuity" at #x=pi#).