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Magazine Chapter 13: Problem 6
Find the limit. $$ \lim _{t \rightarrow \infty}\left\langle t e^{-t}, \frac{t^{3}+t}{2 t^{3}-1}, t \sin \frac{1}{t}\right\rangle $$
Short Answer
Expert verified
The limit is \( \langle 0, \frac{1}{2}, 1 \rangle \).
Step by step solution
01
Analyze Each Component Separately
The problem asks us to find the limit of a vector function as \( t \) approaches infinity. The vector is \( \langle t e^{-t}, \frac{t^{3}+t}{2 t^{3}-1}, t \sin \frac{1}{t} \rangle \). We'll consider each component of the vector separately to find their limits.
02
Find the Limit of the First Component
Consider the first component \( t e^{-t} \). As \( t \to \infty \), the exponential function \( e^{-t} \) tends to zero much faster than the linear function \( t \) grows. Therefore, the limit is \( \lim_{t \to \infty} t e^{-t} = 0 \).
03
Find the Limit of the Second Component
Consider the second component \( \frac{t^{3}+t}{2 t^{3}-1} \). Divide all terms by \( t^3 \) to simplify: \( \frac{1 + \frac{1}{t^2}}{2 - \frac{1}{t^3}} \). As \( t \to \infty \), \( \frac{1}{t^2} \) and \( \frac{1}{t^3} \) approach zero, so the limit is \( \frac{1}{2} \).
04
Find the Limit of the Third Component
Consider the third component \( t \sin \frac{1}{t} \). As \( t \to \infty \), the expression \( \sin \frac{1}{t} \) is approximately \( \frac{1}{t} \), because \( \sin x \approx x \) for small values of \( x \). Thus, \( t \sin \frac{1}{t} \approx t \cdot \frac{1}{t} = 1 \). The limit is \( 1 \).
05
Combine the Limits
Now that we have the limits for each component, we can write the final result as the limit of the original vector: \( \langle 0, \frac{1}{2}, 1 \rangle \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are mathematical expressions that involve an exponent with a variable in the form of \( e^x \), where \( e \) is a constant approximately equal to 2.71828, known as Euler's number.
In the given exercise, we deal with the component \( t e^{-t} \). This requires understanding the behavior of exponential decay as \( t \) approaches infinity.
Key concepts include:
In the given exercise, we deal with the component \( t e^{-t} \). This requires understanding the behavior of exponential decay as \( t \) approaches infinity.
Key concepts include:
- The exponent \( -t \) results in a rapid decrease towards zero for \( e^{-t} \) as \( t \) increases.
- This decay usually outpaces any polynomial growth term like \( t \), which makes expressions like \( t e^{-t} \) approach zero.
Rational Functions
Rational functions are ratios of polynomials represented generally as \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
In our problem, we consider the component \( \frac{t^{3}+t}{2 t^{3}-1} \). To find the limit of rational functions at infinity, we often compare the degrees of the polynomials in the numerator and denominator.
Important points to note:
In our problem, we consider the component \( \frac{t^{3}+t}{2 t^{3}-1} \). To find the limit of rational functions at infinity, we often compare the degrees of the polynomials in the numerator and denominator.
Important points to note:
- Here, both the numerator and the denominator are predominantly influenced by the terms with the highest degree, which is \( t^3 \).
- By dividing every term by \( t^3 \), the expression simplifies, allowing smaller degree terms such as \( \frac{1}{t^2} \) and \( \frac{1}{t^3} \) to become negligible as \( t \rightarrow \infty \).
- The limit is determined by the leading coefficients of these highest degree terms, leading us to \( \frac{1}{2} \).
Sine Function Approximations
Sine functions, like \( \sin x \), are periodic and oscillate between -1 and 1, but their approximations play a crucial role when the argument approaches zero.
In the given exercise, you see \( t \sin \frac{1}{t} \). As \( t \) increases, \( \frac{1}{t} \) becomes very small, meaning we can use an approximation: \( \sin x \approx x \) for small \( x \). Thus, \( \sin \frac{1}{t} \approx \frac{1}{t} \).
Key ideas include:
In the given exercise, you see \( t \sin \frac{1}{t} \). As \( t \) increases, \( \frac{1}{t} \) becomes very small, meaning we can use an approximation: \( \sin x \approx x \) for small \( x \). Thus, \( \sin \frac{1}{t} \approx \frac{1}{t} \).
Key ideas include:
- This approximation helps in simplifying products like \( t \sin \frac{1}{t} \), reducing it to \( t \cdot \frac{1}{t} = 1 \).
- It's effective for finding limits as \( t \to \infty \) because it converts complex transcendental expressions into manageable algebraic forms.
