# Question #2e37e

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: ${\lim}_{x \to 0 -} f \left(x\right) = {\lim}_{x \to 0 -} {e}^{x} = {e}^{0} = 1$.

The right-hand limit as $x$ approaches zero from the right uses the second formula: ${\lim}_{x \to 0 +} f \left(x\right) = {\lim}_{x \to 0 +} \sin \left(x\right) = \sin \left(0\right) = 0$.

Since $1 \ne 0$, the two-sided limit ${\lim}_{x \to 0} f \left(x\right)$ 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: ${\lim}_{x \to \pi -} f \left(x\right) = {\lim}_{x \to \pi -} \sin \left(x\right) = \sin \left(\pi\right) = 0$.

The right-hand limit as $x$ approaches $\pi$ from the right uses the third formula: ${\lim}_{x \to \pi +} f \left(x\right) = {\lim}_{x \to \pi +} \left(x - \pi - 1\right) = \pi - \pi - 1 = - 1$.

Since $0 \ne - 1$, the two-sided limit ${\lim}_{x \to \pi} f \left(x\right)$ does not exist, so $f$ is also not continuous at $x = \pi$ (there is also a "jump discontinuity" at $x = \pi$).