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).