1. **State the problem:** Find the limit as $t \to 0$ of the vector function $$\left\langle \frac{4e^t - 4}{t}, \frac{\sqrt{1+t} - 1}{t}, \frac{5}{1+t} \right\rangle$$ 2. **Recall important formulas and rules:** - The limit of a vector function is the vector of the limits of its components. - Use the definition of the derivative or expansions for limits of indeterminate forms. - For small $t$, $e^t \approx 1 + t + \frac{t^2}{2} + \cdots$ - For small $t$, $\sqrt{1+t} \approx 1 + \frac{t}{2} - \frac{t^2}{8} + \cdots$ 3. **Evaluate the first component:** $$\lim_{t \to 0} \frac{4e^t - 4}{t} = 4 \lim_{t \to 0} \frac{e^t - 1}{t}$$ Using the derivative of $e^t$ at 0: $$\lim_{t \to 0} \frac{e^t - 1}{t} = 1$$ So, $$\lim_{t \to 0} \frac{4e^t - 4}{t} = 4 \times 1 = 4$$ 4. **Evaluate the second component:** $$\lim_{t \to 0} \frac{\sqrt{1+t} - 1}{t}$$ Multiply numerator and denominator by the conjugate: $$\frac{\sqrt{1+t} - 1}{t} \times \frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1} = \frac{1+t - 1}{t(\sqrt{1+t} + 1)} = \frac{t}{t(\sqrt{1+t} + 1)}$$ Cancel $t$: $$\frac{\cancel{t}}{\cancel{t}(\sqrt{1+t} + 1)} = \frac{1}{\sqrt{1+t} + 1}$$ Taking the limit as $t \to 0$: $$\frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$ 5. **Evaluate the third component:** $$\lim_{t \to 0} \frac{5}{1+t} = \frac{5}{1+0} = 5$$ 6. **Combine the results:** $$\lim_{t \to 0} \left\langle \frac{4e^t - 4}{t}, \frac{\sqrt{1+t} - 1}{t}, \frac{5}{1+t} \right\rangle = \langle 4, \frac{1}{2}, 5 \rangle$$