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\( \left. \begin{array} { c } { \lim _ { t \rightarrow 0 } ( t ^ { 2 } \vec { \imath } + 3 t \vec { \jmath } + \frac { 1 - \cos t } { t } \vec { k } ) } \\ { = \langle [ ? ] , [ \quad ] , [ \quad ] \rangle } \\ { } \end{array} \right. \)

Ask by Washington Cross. in the United States
Feb 03,2025

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Tutor-Verified Answer

Answer

\[ \left\langle 0, \; 0, \; 0 \right\rangle \]

Solution

To evaluate the limit: \[ \lim_{t \rightarrow 0} \left( t^2 \vec{\imath} + 3t \vec{\jmath} + \frac{1 - \cos t}{t} \vec{k} \right) \] we can examine each component separately. 1. **\(\vec{\imath}\) component:** \[ \lim_{t \rightarrow 0} t^2 = 0 \] 2. **\(\vec{\jmath}\) component:** \[ \lim_{t \rightarrow 0} 3t = 0 \] 3. **\(\vec{k}\) component:** \[ \lim_{t \rightarrow 0} \frac{1 - \cos t}{t} \] Using the Taylor series expansion for \(\cos t\) around \(t = 0\): \[ \cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots \] Therefore: \[ 1 - \cos t = \frac{t^2}{2} - \frac{t^4}{24} + \cdots \] Dividing by \(t\): \[ \frac{1 - \cos t}{t} = \frac{t}{2} - \frac{t^3}{24} + \cdots \] Taking the limit as \(t \rightarrow 0\): \[ \lim_{t \rightarrow 0} \frac{1 - \cos t}{t} = 0 \] Putting it all together, the limit vector is: \[ \left\langle 0, \; 0, \; 0 \right\rangle \] So, filling in the blanks: \[ \left. \begin{array}{c} \lim_{t \rightarrow 0} \left( t^2 \vec{\imath} + 3t \vec{\jmath} + \frac{1 - \cos t}{t} \vec{k} \right) \\ = \langle\, 0 \,,\, 0 \,,\, 0 \,\rangle \end{array} \right. \]

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As \( t \) approaches 0, we can evaluate each component of the vector separately. The x-component is simply \( \lim_{t \to 0} t^2 = 0 \). For the y-component, \( \lim_{t \to 0} 3t = 0 \) as well. The z-component involves a special limit: \( \lim_{t \to 0} \frac{1 - \cos t}{t} \). Using L'Hôpital's Rule or recognizing that \( 1 - \cos t \approx \frac{t^2}{2} \) near 0 yields \( \lim_{t \to 0} \frac{1 - \cos t}{t} = 0 \). Putting it all together, we have \( \langle 0, 0, 0 \rangle \).

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