Questions: For problems 1-6 evaluate the given limit. 1. lim t -> 0 (cos(2t)i - e^(4-t)j + (t^2 + 3t - 9)k) 2. lim t -> 4 <(t-4)/(t^2 - 3t - 4), (t^2 - 4t)/(16 - t^2)> 3. lim t -> 0 <sin(t)/(2t), (1 - cos(t))/t, -3> 4. lim t -> -8 ((e^(t^2 - 64) - 1)/(t + 8)i + sin(t + 8)/(t + 8)j - k) 5. lim t -> -∞ <(5t^2 - 8t + 1)/(12 + 5t^2), (2 + t^3)/(1 + t^2 + t^4)> 6. lim t -> ∞ <ln(1 - 4/t), e^(1/t^2), 2>

Solution Steps

1. For the first limit, evaluate each component of the vector separately as \( t \) approaches 0. The cosine and exponential functions are continuous, so substitute \( t = 0 \) directly. For the polynomial, substitute \( t = 0 \) as well.

2. For the second limit, simplify each component of the vector by factoring and canceling terms where possible. Then, substitute \( t = 4 \) into the simplified expressions.

3. For the third limit, use L'Hôpital's Rule for the indeterminate forms in the first two components. The third component is constant, so it remains unchanged.

We need to evaluate the limit: \[ \lim_{t \rightarrow 0}\left(\cos(2t) \vec{i} - e^{4-t} \vec{j} + (t^2 + 3t - 9) \vec{k}\right) \] Calculating each component as \( t \) approaches 0:

• For the \( \vec{i} \) component: \[ \lim_{t \rightarrow 0} \cos(2t) = \cos(0) = 1 \]
• For the \( \vec{j} \) component: \[ \lim_{t \rightarrow 0} e^{4-t} = e^{4} \]
• For the \( \vec{k} \) component: \[ \lim_{t \rightarrow 0} (t^2 + 3t - 9) = 0 + 0 - 9 = -9 \] Thus, the limit is: \[ \vec{L_1} = \begin{pmatrix} 1 \\ -e^{4} \\ -9 \end{pmatrix} \]

We need to evaluate the limit: \[ \lim_{t \rightarrow 4}\left\langle \frac{t-4}{t^2 - 3t - 4}, \frac{t^2 - 4t}{16 - t^2} \right\rangle \] Calculating each component:

• For the first component: \[ \lim_{t \rightarrow 4} \frac{t-4}{t^2 - 3t - 4} = \frac{0}{(4)^2 - 3(4) - 4} = \frac{0}{0} \text{ (indeterminate)} \] After simplification, we find: \[ \lim_{t \rightarrow 4} \frac{1}{5} = \frac{1}{5} \]
• For the second component: \[ \lim_{t \rightarrow 4} \frac{t^2 - 4t}{16 - t^2} = \frac{0}{0} \text{ (indeterminate)} \] After simplification, we find: \[ \lim_{t \rightarrow 4} -\frac{1}{2} = -\frac{1}{2} \] Thus, the limit is: \[ \vec{L_2} = \begin{pmatrix} \frac{1}{5} \\ -\frac{1}{2} \end{pmatrix} \]

We need to evaluate the limit: \[ \lim_{t \rightarrow 0}\left\langle \frac{\sin(t)}{2t}, \frac{1 - \cos(t)}{t}, -3 \right\rangle \] Calculating each component:

• For the first component: \[ \lim_{t \rightarrow 0} \frac{\sin(t)}{2t} = \frac{1}{2} \]
• For the second component: \[ \lim_{t \rightarrow 0} \frac{1 - \cos(t)}{t} = 0 \]
• The third component is constant: \[ -3 \] Thus, the limit is: \[ \vec{L_3} = \begin{pmatrix} \frac{1}{2} \\ 0 \\ -3 \end{pmatrix} \]

Final Answer