Answer
a) $t \in (-1,0) \cup (0,2]$
b) $\lim\limits_{t \to 0} r(t)=\lt \sqrt 2, 1, 0 \gt$
c) $r'(t) =\lt \dfrac{-1}{2\sqrt {2-t}},\dfrac{te^t-e^t+1}{t^2}, \dfrac{1}{t+1} \gt$
Work Step by Step
a) Domain of $x$ component: $\sqrt{2-t}$ domain which is $t \in (-\infty, 2]$
Also,
Domain of $y$ component: $\dfrac{e^t-1}{t}$ domain which is $t \in (-\infty,0) \cup (0, \infty)$
Domain of $z$ component: $ln(1+t)$ domain which is $t \in (-1, \infty)$
Thus, we get $t \in (-1,0) \cup (0,2]$
b) Here, $\lim\limits_{t \to 0}\sqrt {2-t}=\sqrt {2-0}=\sqrt 2$
$\lim\limits_{t \to 0} =\lim\limits_{t \to 0} \dfrac{e^t}{t}=1\\ \lim\limits_{t \to 0} \ln(1+t)=\ln (1+0)=0$
Hence, we get $\lim\limits_{t \to 0} r(t)=\lt \sqrt 2, 1, 0 \gt$
c) After taking the derivative of each component we get
$r'(t) =\lt \dfrac{-1}{2\sqrt {2-t}},\dfrac{te^t-e^t+1}{t^2}, \dfrac{1}{t+1} \gt$
