Difference between revisions of "Limits of Vector Functions"

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http://mathonline.wikidot.com/limits-of-vector-valued-functions
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'''Definition:''' If <span class="math-inline"><math>\vec{r}(t) = (x(t), y(t), z(t))</math></span> is a vector-valued function, then <span class="math-inline"><math>\lim_{t \to a} \vec{r}(t) = \left ( \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t) \right )</math></span> provided that the limits of the components exist.
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//Limits of vector-valued functions in <math>\mathbb{R}^n</math> are defined similarly as the limit of each component. //
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Let's look at some examples of evaluating limits of vector-valued functions. Consider the vector-valued function <math>\vec{r}(t) = (t^2 - 1, t + 1, e^t)</math> and suppose that we wanted to compute <math>\lim_{t \to 2} \vec{r}(t)</math>. To compute this limit, all we need to do is compute the limits of the components.
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<div style="text-align: center;"> <math>\begin{align} \quad \lim_{t \to 2} \vec{r}(t) = \left ( \lim_{t \to 2} t^2 - 1, \lim_{t \to 2} t + 1, \lim_{t \to 2} e^t \right ) = (3, 3, e^2) \end{align}</math></div>
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For another example, consider the vector-valued function <math>\vec{r}(t) = \left ( \frac{e^t - 1}{t}, \frac{t - 1}{t+1}, t^2 + 3 \right )</math> and suppose that we wanted to compute <math>\lim_{t \to 0} \vec{r}(t)</math>. To compute this limit, we will compute all of the limits of the components again, however, this time the limits are a little trickier to compute. Fortunately, we have already learned about various rules to evaluate limits.
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<div style="text-align: center;"><math>\begin{align} \quad \lim_{t \to 0} \vec{r}(t) = \left ( \lim_{t \to 0} \frac{e^t - 1}{t}, \lim_{t \to 0} \frac{t - 1}{t + 1}, \lim_{t \to 0} t^2 + 3 \right ) \end{align}</math></div>
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For <math>\lim_{t \to 0} \frac{e^t - 1}{t}</math> we will use L'Hospital's Rule, and so <math>\lim_{t \to 0} \frac{e^t - 1}{t} \overset{H} = \lim_{t \to 0} \frac{e^t}{1} = 1</math>.
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For <math>\lim_{t \to 0} \frac{t - 1}{t+1}</math>, we can use direct substitution and so <math>\lim_{t \to 0} \frac{t-1}{t+1} = -1</math>.
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Now <math>\lim_{t \to 0} t^2 + 3</math> is also easy to compute by direct substitution and so <math>\lim_{t \to 0} t^2 + 3 = 3</math>.
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Thus we have that <math>\lim_{t \to 0} \vec{r}(t) = (1, -1, 3)</math>.
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The following theorem gives us a formal definition to say a vector-valued function <math>\vec{r}(t)</math> has limit <math>\vec{b}</math> at <math>t = a</math>, which is analogous to that of limits of real-valued functions.
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:'''Theorem 1:''' Let <math>\vec{r}(t) = (x(t), y(t), z(t))</math> be a vector-valued function and let <math>\vec{b} = (b_1, b_2, b_3) \in \mathbb{R}^3</math>. Then <math>\lim_{t \to a} \vec{r}(t) = \vec{b}</math> if and only if <math>\forall \epsilon > 0</math> <math>\exists \delta > 0</math> such that if <math>0 < \mid t - a \mid < \delta</math> then <math>\| \vec{r}(t) - \vec{b} \| < \epsilon</math>.
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* '''Proof:''' <math>\Rightarrow</math> Suppose that <math>\lim_{t \to a} \vec{r}(t) = \vec{b}</math>. Then we have that:</li>
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<div style="text-align: center;"><math>\begin{align} \left ( \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t) \right ) = (b_1, b_2, b_3) \end{align}</math></div>
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* Now recall that two vectors are equal if and only if their components are equal, and so the equation above implies that <math>\lim_{t \to a} x(t) = b_1</math>, <math>\lim_{t \to a} y(t) = b_2</math>, and <math>\lim_{t \to a} z(t) = b_3</math>. Now notice that these three limits are limits of real-valued functions.
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* Since <math>\lim_{t \to a} x(t) = b_1</math> then <math>\forall \epsilon > 0</math> <math>\exists \delta_1 > 0</math> such that if <math>0 < \mid t - a \mid < \delta_1</math> then <math>\mid x(t) - b_1 \mid < \frac{\epsilon}{3}</math>.
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* Since <math>\lim_{t \to a} y(t) = b_2</math> then <math>\forall \epsilon > 0</math> <math>\exists \delta_2 > 0</math> such that if <math>0 < \mid t - a \mid < \delta_2</math> then <math>\mid y(t) - b_2 \mid < \frac{\epsilon}{3}</math>.
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* Since <math>\lim_{t \to a} z(t) = b_3</math> then <math>\forall \epsilon > 0</math> <math>\exists \delta_3 > 0</math> such that if <math>0 < \mid t - a \mid < \delta_3</math> then <math>\mid z(t) - b_3 \mid < \frac{\epsilon}{3}</math>.
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* Let <math>\delta = \mathrm{min} \{ \delta_1, \delta_2, \delta_3 \}</math>. Then if <math>0 < \mid t - a \mid < \delta</math> we have that:
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<div style="text-align: center;"><math>\begin{align} \quad \quad \| \vec{r}(t) - \vec{b} \| = \| (x(t) - b_1, y(t) - b_2, z(t) - b_3)) \| = \sqrt{(x(t) - b_1)^2 + (y(t) - b_2)^2 + (z(t) - b_3)^2} \\ \quad \quad \leq \sqrt{(x(t) - b_1)^2} + \sqrt{(y(t) - b_2)^2} + \sqrt{(z(t) - b_3)^2} = \mid x(t) - b_1 \mid + \mid y(t) - b_2 \mid + \mid z(t) - b_3 \mid < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{align}</math></div>
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* <math>\Leftarrow</math> Suppose that <math>\forall \epsilon > 0</math> <math>\exists \delta > 0</math> such that if <math>0 < \mid t - a \mid < \delta</math> then <math>\| \vec{r}(t) - \vec{b} \| < \epsilon</math>. Therefore we have that
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:<math>\| \vec{r}(t) - \vec{b} \| = \| (x(t) - b_1, y(t) - b_2, z(t) - b_3) \| < \epsilon</math>, which implies that:
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<div style="text-align: center;"><math>\begin{align} \sqrt{(x(t) - b_1)^2 + (y(t) - b_2)^2 + (z(t) - b_3)^2} < \epsilon \\ (x(t) - b_1)^2 + (y(t) - b_2)^2 + (z(t) - b_3)^2 < \epsilon^2 \\ \end{align}</math></div>
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* Now since all terms of the lefthand side of this equation are positive, we must have that for <math>0 < \mid t - a \mid < \delta</math> then <math>(x(t) - b_1)^2 < \epsilon^2</math>, <math>(y(t) - b_2)^2 < \epsilon^2</math>, and <math>(z(t) - b_3)^2 < \epsilon^2</math>, and so <math>\mid x(t) - b_1 \mid < \epsilon</math>, <math>\mid y(t) - b_2 \mid < \epsilon</math> and <math>\mid z(t) - b_3 \mid < \epsilon</math>. Therefore by the definition of real-valued function limits we have that <math>\lim_{t \to a} x(t) = b_1</math>, <math>\lim_{t \to a} y(t) = b_2</math>, and <math>\lim_{t \to a} z(t) = b_3</math>.
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* Thus <math>\lim_{t \to a} \vec{r}(t) = \left ( \lim_{t \to a} x(t), \lim_{t \to a} y(t), \lim_{t \to a} z(t) \right ) = (b_1, b_2, b_3) = \vec{b}</math>. <math>\blacksquare</math></li>
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== Licensing ==
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Content obtained and/or adapted from:
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* [http://mathonline.wikidot.com/limits-of-vector-valued-functions Limits of Vector-Valued Functions, mathonline.wikidot.com] under a CC BY-SA license

Latest revision as of 16:56, 29 October 2021

Definition: If is a vector-valued function, then provided that the limits of the components exist.

//Limits of vector-valued functions in are defined similarly as the limit of each component. //

Let's look at some examples of evaluating limits of vector-valued functions. Consider the vector-valued function and suppose that we wanted to compute . To compute this limit, all we need to do is compute the limits of the components.

For another example, consider the vector-valued function and suppose that we wanted to compute . To compute this limit, we will compute all of the limits of the components again, however, this time the limits are a little trickier to compute. Fortunately, we have already learned about various rules to evaluate limits.

For we will use L'Hospital's Rule, and so .

For , we can use direct substitution and so .

Now is also easy to compute by direct substitution and so .

Thus we have that .

The following theorem gives us a formal definition to say a vector-valued function has limit at , which is analogous to that of limits of real-valued functions.

Theorem 1: Let be a vector-valued function and let . Then if and only if such that if then .
  • Proof: Suppose that . Then we have that:
  • Now recall that two vectors are equal if and only if their components are equal, and so the equation above implies that , , and . Now notice that these three limits are limits of real-valued functions.
  • Since then such that if then .
  • Since then such that if then .
  • Since then such that if then .
  • Let . Then if we have that:
  • Suppose that such that if then . Therefore we have that
, which implies that:


  • Now since all terms of the lefthand side of this equation are positive, we must have that for then , , and , and so , and . Therefore by the definition of real-valued function limits we have that , , and .
  • Thus .

Licensing

Content obtained and/or adapted from: