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Section 12.7 : Calculus with Vector Functions
2. Evaluate the following limit.
\[\mathop {\lim }\limits_{t \to - 2} \left( {\frac{{1 - {{\bf{e}}^{t + 2}}}}{{{t^2} + t - 2}}\vec i + \vec j + \left( {{t^2} + 6t} \right)\vec k} \right)\] Show SolutionThere really isn’t a lot to do here with this problem. All we need to do is take the limit of all the components of the vector.
\[\begin{align*}\mathop {\lim }\limits_{t \to - 2} \left( {\frac{{1 - {{\bf{e}}^{t + 2}}}}{{{t^2} + t - 2}}\vec i + \vec j + \left( {{t^2} + 6t} \right)\vec k} \right) & = \mathop {\lim }\limits_{t \to - 2} \frac{{1 - {{\bf{e}}^{t + 2}}}}{{{t^2} + t - 2}}\vec i + \mathop {\lim }\limits_{t \to - 2} \vec j + \mathop {\lim }\limits_{t \to - 2} \left( {{t^2} + 6t} \right)\vec k\\ & = \mathop {\lim }\limits_{t \to - 2} \frac{{ - {{\bf{e}}^{t + 2}}}}{{2t + 1}}\vec i + \mathop {\lim }\limits_{t \to - 2} \vec j + \mathop {\lim }\limits_{t \to - 2} \left( {{t^2} + 6t} \right)\vec k = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}\vec i + \vec j - 8\vec k}}\end{align*}\]Don’t forget L’Hospital’s Rule! We needed that for the first term.