I am given the function
$\gamma:[-1,\frac{\pi}{2}] \rightarrow \mathbb{C}$
$\gamma(t) =
\begin{cases}
t+1 & \text{for $-1 \leq t \leq0$} \\
e^{it} & \text{for $0 \leq t \leq\frac{\pi}{2}$} \\
\end{cases}$
And I want to show this function is continuous and piecewise continuously differentiable. To show the function is continuous at the endpoints $-1$ and $\frac{\pi}{2}$ of the interval $[-1,\frac{\pi}{2}]$, is it sufficient to show that $\lim_{t\rightarrow-1^{+}} \gamma(t) = \gamma(-1)$ and $\lim_{t\rightarrow\frac{\pi}{2}^{-}} \gamma(t) = \gamma(\frac{\pi}{2})$ ?
Then, to show the function is piecewise differentiable, would I need to show that $\gamma_{|[-1,0]}$ is differentiable at all interior points of the interval $[-1,0]$, left differentiable at $0$ and right differentiable at $-1$? (And then do the equivalent for $\gamma_{|[0,\frac{\pi}{2}]}$)
To show continuity at the endpoints of the interval for this derivative, would I need to show that $\lim_{t\rightarrow-1^{+}} \gamma_{|[-1,0]}'(t) = \gamma_{|[-1,0]}'(-1^{+})$ and $\lim_{t\rightarrow0^{-}} \gamma_{|[-1,0]}'(t) = \gamma_{|[-1,0]}'(0^{-})$? (And then do the equivalent for $\gamma_{|[0,\frac{\pi}{2}]}$)