1. **State the problem:** We are given the vector function $$r(t) = \left(\tan(t) \cot(t), \ln t, \sqrt{\pi^2 - t^2}\right)$$ and need to find its domain and the limit $$\lim_{t \to 1} r(t).$$ 2. **Find the domain:** The domain is the set of all $t$ values for which each component of $r(t)$ is defined. - For $\tan(t) \cot(t)$: Both $\tan(t)$ and $\cot(t)$ must be defined. $\tan(t)$ is undefined at $t = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$. $\cot(t)$ is undefined at $t = k\pi$, $k \in \mathbb{Z}$. So $t$ cannot be any integer multiple of $\pi$ or $\frac{\pi}{2} + k\pi$. - For $\ln t$: $t > 0$. - For $\sqrt{\pi^2 - t^2}$: The radicand must be non-negative, so $$\pi^2 - t^2 \geq 0 \implies |t| \leq \pi.$$ Combining with $t > 0$, we get $$0 < t \leq \pi.$$ - Combining all restrictions, $t$ must be in $(0, \pi]$ excluding points where $\tan(t)$ or $\cot(t)$ are undefined, i.e., excluding $t = \frac{\pi}{2}$ and $t = \pi$ (since $\cot(\pi)$ is undefined). But $t=\pi$ is at the boundary and $\cot(\pi)$ is undefined, so exclude $t=\pi$. Thus, domain is $$\{ t \in (0, \pi) \mid t \neq \frac{\pi}{2} \}.$$ 3. **Simplify the first component:** Recall that $$\cot(t) = \frac{1}{\tan(t)}$$ for $t$ where both are defined. So, $$\tan(t) \cot(t) = \tan(t) \times \frac{1}{\tan(t)} = 1,$$ except at points where $\tan(t)$ or $\cot(t)$ are undefined. 4. **Evaluate the limit:** $$\lim_{t \to 1} r(t) = \lim_{t \to 1} \left(1, \ln t, \sqrt{\pi^2 - t^2}\right) = \left(1, \ln 1, \sqrt{\pi^2 - 1^2}\right).$$ Since $\ln 1 = 0$ and $\sqrt{\pi^2 - 1} = \sqrt{\pi^2 - 1}$, $$\lim_{t \to 1} r(t) = \left(1, 0, \sqrt{\pi^2 - 1}\right).$$ **Final answers:** - Domain: $$\{ t \in (0, \pi) \mid t \neq \frac{\pi}{2} \}$$ - Limit: $$\lim_{t \to 1} r(t) = \left(1, 0, \sqrt{\pi^2 - 1}\right).$$