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Section 12.7 : Calculus with Vector Functions
3. Evaluate the following limit.
\[\mathop {\lim }\limits_{t \to \infty } \left\langle {\frac{1}{{{t^2}}},\frac{{2{t^2}}}{{1 - t - {t^2}}},{{\bf{e}}^{ - t}}} \right\rangle \] 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.
\[\begin{align*}\mathop {\lim }\limits_{t \to \infty } \left\langle {\frac{1}{{{t^2}}},\frac{{2{t^2}}}{{1 - t - {t^2}}},{{\bf{e}}^{ - t}}} \right\rangle & = \left\langle {\mathop {\lim }\limits_{t \to \infty } \frac{1}{{{t^2}}},\mathop {\lim }\limits_{t \to \infty } \frac{{2{t^2}}}{{1 - t - {t^2}}},\mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{ - t}}} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to \infty } \frac{1}{{{t^2}}},\mathop {\lim }\limits_{t \to \infty } \frac{{2{t^2}}}{{{t^2}\left( {\frac{1}{{{t^2}}} - \frac{1}{t} - 1} \right)}},\mathop {\lim }\limits_{t \to \infty } {{\bf{e}}^{ - t}}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {0, - 2,0} \right\rangle }}\end{align*}\]Don’t forget your basic limit at infinity processes/facts.