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) Here, the domain of the $x$ component is:
$\sqrt{2-t}$
domain is $t \in (-\infty, 2]$
The domain of the $y$ component is:
$\dfrac{e^t-1}{t}$
domain is $t \in (-\infty,0) \cup (0, \infty)$
The domain of the $z$ component is:
$ln(1+t)$
domain is $t \in (-1, \infty)$
Hence, $t \in (-1,0) \cup (0,2]$
b) Here, $\lim\limits_{t \to 0}\sqrt {2-t}=\sqrt 2$
$\lim\limits_{t \to 0} =\lim\limits_{t \to 0} \dfrac{e^t}{t}=1$
and $ \lim\limits_{t \to 0}ln(1+t)=\ln 1=0$
Hence, $\lim\limits_{t \to 0} r(t)=\lt \sqrt 2, 1, 0 \gt$
c) We need to take the derivative of each component.
Here, $r'(t) =\lt \dfrac{-1}{2\sqrt {2-t}},\dfrac{te^t-e^t+1}{t^2}, \dfrac{1}{t+1} \gt$
