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Section 12.7 : Calculus with Vector Functions
1. Evaluate the following limit.
\[\mathop {\lim }\limits_{t \to 1} \left\langle {{{\bf{e}}^{t - 1}},4t,\frac{{t - 1}}{{{t^2} - 1}}} \right\rangle \] 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.
\[\begin{align*}\mathop {\lim }\limits_{t \to 1} \left\langle {{{\bf{e}}^{t - 1}},4t,\frac{{t - 1}}{{{t^2} - 1}}} \right\rangle & = \left\langle {\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{t - 1}},\mathop {\lim }\limits_{t \to 1} 4t,\mathop {\lim }\limits_{t \to 1} \frac{{t - 1}}{{{t^2} - 1}}} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{t - 1}},\mathop {\lim }\limits_{t \to 1} 4t,\mathop {\lim }\limits_{t \to 1} \frac{1}{{2t}}} \right\rangle = \left\langle {{{\bf{e}}^0},4,\frac{1}{2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {1,4,\frac{1}{2}} \right\rangle }}\end{align*}\]Don’t forget L’Hospital’s Rule! We needed that for the third term.