Paul's Online Notes
Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Calculus with Vector Functions
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

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 Solution

There 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.