We are asked to find the limit of a vector-valued function as $t$ approaches infinity. The vector-valued function is given by: $\lim_{t \to \infty} (\frac{t^2+1}{3t^2-2} \hat{i} + \frac{1}{t} \hat{j} + \frac{1}{t^2+2} \hat{k})$

1. Problem Description

We are asked to find the limit of a vector-valued function as $t$ approaches infinity. The vector-valued function is given by:

$\lim_{t \to \infty} (\frac{t^2+1}{3t^2-2} \hat{i} + \frac{1}{t} \hat{j} + \frac{1}{t^2+2} \hat{k})$

2. Solution Steps

To find 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 $\hat{i}$ component:

$\lim_{t \to \infty} \frac{t^2+1}{3t^2-2}$

To evaluate this limit, we can divide both the numerator and denominator by the highest power of $t$, which is $t^2$:

$\lim_{t \to \infty} \frac{\frac{t^2}{t^2}+\frac{1}{t^2}}{\frac{3t^2}{t^2}-\frac{2}{t^2}} = \lim_{t \to \infty} \frac{1+\frac{1}{t^2}}{3-\frac{2}{t^2}}$

As $t \to \infty$, $\frac{1}{t^2} \to 0$ and $\frac{2}{t^2} \to 0$.

Therefore, $\lim_{t \to \infty} \frac{1+\frac{1}{t^2}}{3-\frac{2}{t^2}} = \frac{1+0}{3-0} = \frac{1}{3}$.

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

$\lim_{t \to \infty} \frac{1}{t} = 0$.

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

$\lim_{t \to \infty} \frac{1}{t^2+2}$

As $t \to \infty$, $t^2 \to \infty$, and $t^2 + 2 \to \infty$.

Therefore, $\lim_{t \to \infty} \frac{1}{t^2+2} = 0$.

Thus, the limit of the vector-valued function is:

$\frac{1}{3} \hat{i} + 0 \hat{j} + 0 \hat{k} = \frac{1}{3} \hat{i}$

3. Final Answer

$\frac{1}{3} \hat{i}$