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We are asked to compute the limit of the vector \(\left\langle \arctan(t), e^{-3t}, \frac{\ln(6t)}{t} \right\rangle\) as \(t \to \infty\).

Let's break this down component-wise:

### 1. \(\lim_{t \to \infty} \arctan(t)\):
- The arctangent function, \(\arctan(t)\), has a horizontal asymptote at \(\frac{\pi}{2}\) as \(t\) approaches infinity.
- Therefore, \[
\lim_{t \to \infty} \arctan(t) = \frac{\pi}{2}.
\]

### 2. \(\lim_{t \to \infty} e^{-3t}\):
- The exponential function decays to zero as \(t\) approaches infinity since \(e^{-3t}\) becomes smaller and smaller.
- Therefore, \[
\lim_{t \to \infty} e^{-3t} = 0.
\]

### 3. \(\lim_{t \to \infty} \frac{\ln(6t)}{t}\):
- This is a logarithmic function divided by a linear term. We can use L'Hopital's Rule because this is an indeterminate form of type \(\frac{\infty}{\infty}\).
- Differentiate the numerator and denominator:
  - The derivative of \(\ln(6t)\) is \(\frac{1}{t}\),
  - The derivative of \(t\) is \(1\).
- Applying L'Hopital's Rule:
  \[
  \lim_{t \to \infty} \frac{\ln(6t)}{t} = \lim_{t \to \infty} \frac{\frac{1}{t}}{1} = \lim_{t \to \infty} \frac{1}{t} = 0.
  \]

### Final Answer:
The limit of the vector is \[
\lim_{t \to \infty} \left\langle \arctan(t), e^{-3t}, \frac{\ln(6t)}{t} \right\rangle = \left\langle \frac{\pi}{2}, 0, 0 \right\rangle.
\]

Would you like any further details or clarification on any part?

Here are some related questions you might find helpful:
1. How does L'Hopital's Rule work, and when can it be applied?
2. What is the behavior of exponential functions as \(t \to \infty\)?
3. Can arctangent functions have asymptotes?
4. How do logarithmic functions behave compared to linear and polynomial functions at infinity?
5. What are the common indeterminate forms that can occur in limits?

**Tip:** When evaluating limits of logarithmic and polynomial functions, comparing growth rates (like using L'Hopital's Rule) is very useful.

Find the limit: lim t → ∞ ⟨arctan(t), e^(-3t), ln(6t)/t⟩

This page covers the limit problem: lim t → ∞ ⟨arctan(t), e^(-3t), ln(6t)/t⟩. It includes a step-by-step solution, utilizing L'Hopital's Rule for indeterminate forms, and examines the behavior of trigonometric and exponential functions as t approaches infinity.

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