We are asked to evaluate the limit of a vector-valued function as $t$ approaches 0. The vector-valued function is given by $$ \lim_{t \to 0} \left[ \frac{\sin t \cos t}{t} \mathbf{i} - \frac{7t^3}{e^t} \mathbf{j} + \frac{t}{t+1} \mathbf{k} \right] $$

1. Problem Description

We are asked to evaluate the limit of a vector-valued function as $t$ approaches

0. The vector-valued function is given by

$$\lim_{t \to 0} \left[ \frac{\sin t \cos t}{t} \mathbf{i} - \frac{7t^3}{e^t} \mathbf{j} + \frac{t}{t+1} \mathbf{k} \right]$$

2. Solution Steps

To evaluate the limit of the vector-valued function, we need to find the limit of each component separately.

First, let's find the limit of the $\mathbf{i}$ component:

$$\lim_{t \to 0} \frac{\sin t \cos t}{t}$$

We can rewrite this as

$$\lim_{t \to 0} \frac{\sin t}{t} \cdot \lim_{t \to 0} \cos t$$

We know that $\lim_{t \to 0} \frac{\sin t}{t} = 1$ and $\lim_{t \to 0} \cos t = \cos(0) = 1$.

Therefore,

$$\lim_{t \to 0} \frac{\sin t \cos t}{t} = 1 \cdot 1 = 1$$

Next, let's find the limit of the $\mathbf{j}$ component:

$$\lim_{t \to 0} \frac{7t^3}{e^t}$$

Since both the numerator and denominator are continuous functions, we can simply plug in $t=0$:

$$\lim_{t \to 0} \frac{7t^3}{e^t} = \frac{7(0)^3}{e^0} = \frac{0}{1} = 0$$

Finally, let's find the limit of the $\mathbf{k}$ component:

$$\lim_{t \to 0} \frac{t}{t+1}$$

Since both the numerator and denominator are continuous functions, we can simply plug in $t=0$:

$$\lim_{t \to 0} \frac{t}{t+1} = \frac{0}{0+1} = \frac{0}{1} = 0$$

So, the limit of the vector-valued function is given by:

$$\lim_{t \to 0} \left[ \frac{\sin t \cos t}{t} \mathbf{i} - \frac{7t^3}{e^t} \mathbf{j} + \frac{t}{t+1} \mathbf{k} \right] = 1\mathbf{i} - 0\mathbf{j} + 0\mathbf{k} = \mathbf{i}$$

3. Final Answer

$\mathbf{i}$