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\( \left. \begin{array} { c } { \lim _ { t \rightarrow \infty } ( e ^ { - t } \vec { \imath } + \frac { 1 } { t } \vec { \jmath } + \frac { t } { t ^ { 2 } + 1 } \vec { k } ) } \\ { = \langle [ ? ] , [ \quad ] , [ \quad ] \rangle } \\ { } \end{array} \right. \)

Ask by Stanley Li. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The limit of the vector as \( t \) approaches infinity is \( \langle 0, \, 0, \, 0 \rangle \).

Solution

To evaluate the limit of the vector as \( t \) approaches infinity: \[ \lim_{t \to \infty} \left( e^{-t} \vec{\imath} + \frac{1}{t} \vec{\jmath} + \frac{t}{t^{2} + 1} \vec{k} \right) \] Let's examine each component individually: 1. **i-component**: \[ \lim_{t \to \infty} e^{-t} = 0 \] As \( t \) approaches infinity, \( e^{-t} \) decays exponentially to 0. 2. **j-component**: \[ \lim_{t \to \infty} \frac{1}{t} = 0 \] As \( t \) grows larger, \( \frac{1}{t} \) approaches 0. 3. **k-component**: \[ \lim_{t \to \infty} \frac{t}{t^{2} + 1} = \lim_{t \to \infty} \frac{1}{t + \frac{1}{t}} = 0 \] Here, dividing numerator and denominator by \( t \) simplifies the expression, showing that it tends to 0 as \( t \) becomes large. Combining these results, the limit vector is: \[ \lim_{t \to \infty} \left( e^{-t} \vec{\imath} + \frac{1}{t} \vec{\jmath} + \frac{t}{t^{2} + 1} \vec{k} \right) = \langle 0, \, 0, \, 0 \rangle \] **Final Answer:** \[ \lim_{t \to \infty} \left( e^{-t} \vec{\imath} + \frac{1}{t} \vec{\jmath} + \frac{t}{t^{2} + 1} \vec{k} \right) = \langle 0, \, 0, \, 0 \rangle \]

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The Deep Dive

As \( t \) approaches infinity, each component of the vector expression behaves differently. The first component, \( e^{-t} \), approaches 0 rapidly because the exponential decay is very strong. The second component, \( \frac{1}{t} \), also trends towards 0 as \( t \) increases. The third component, \( \frac{t}{t^2 + 1} \), simplifies to \( \frac{1}{t} \) for large \( t \), which also tends towards 0. Combining these evaluations, we find that: \[ \lim_{t \rightarrow \infty} \left( e^{-t} \vec{\imath} + \frac{1}{t} \vec{\jmath} + \frac{t}{t^2 + 1} \vec{k} \right) = \langle 0, 0, 0 \rangle \]

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