Math Problem Statement

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

Solution

We are asked to compute the limit of the vector arctan(t),e3t,ln(6t)t\left\langle \arctan(t), e^{-3t}, \frac{\ln(6t)}{t} \right\rangle as tt \to \infty.

Let's break this down component-wise:

1. limtarctan(t)\lim_{t \to \infty} \arctan(t):

  • The arctangent function, arctan(t)\arctan(t), has a horizontal asymptote at π2\frac{\pi}{2} as tt approaches infinity.
  • Therefore, limtarctan(t)=π2.\lim_{t \to \infty} \arctan(t) = \frac{\pi}{2}.

2. limte3t\lim_{t \to \infty} e^{-3t}:

  • The exponential function decays to zero as tt approaches infinity since e3te^{-3t} becomes smaller and smaller.
  • Therefore, limte3t=0.\lim_{t \to \infty} e^{-3t} = 0.

3. limtln(6t)t\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)\ln(6t) is 1t\frac{1}{t},
    • The derivative of tt is 11.
  • Applying L'Hopital's Rule: limtln(6t)t=limt1t1=limt1t=0.\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 limtarctan(t),e3t,ln(6t)t=π2,0,0.\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 tt \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.

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Math Problem Analysis

Mathematical Concepts

Limits
Logarithmic Functions
Exponential Decay
Trigonometric Limits

Formulas

lim t → ∞ arctan(t) = π/2
lim t → ∞ e^(-3t) = 0
lim t → ∞ ln(6t)/t = 0

Theorems

L'Hopital's Rule
Limit of Exponential Functions at Infinity
Trigonometric Limit (arctangent function)

Suitable Grade Level

Undergraduate Level