Question

Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that $$\begin{aligned} &\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 . \text { Prove that }\\\ &\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \text { and } \lim _{t \rightarrow a} h(t)=L_{3}. \end{aligned}$$ $$\begin{aligned} &\text { b. Assume that } \lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text { and }\\\ &\lim _{t \rightarrow a} h(t)=L_{3} . \text { Prove that } \lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3},\right\rangle\\\ &\text { which means that } \lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0. \end{aligned}$$

Short answer

Question: Prove the following statements for a vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) and its limit \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\), where \(\mathbf{L}=\langle L_1, L_2, L_3 \rangle\): a. If \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\), then \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text{ and } \lim _{t \rightarrow a} h(t)=L_{3}\). b. If \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text{ and } \lim _{t \rightarrow a} h(t)=L_{3}\), then \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\).

Step-by-step solution

Writing the definition of the limit

Given \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\), this means that for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that \(|\mathbf{r}(t)-\mathbf{L}| < \epsilon\) when \(0 < |t-a| < \delta\).

Analyzing the component functions

We need to show that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text{ and } \lim _{t \rightarrow a} h(t)=L_{3}\). We have \(\mathbf{r}(t) - \mathbf{L} = \langle f(t)-L_1, g(t)-L_2, h(t)-L_3 \rangle\). Then, we use the inequality \(||\mathbf{A}|| \ge |A_i|\) for any component \(i\) of \(\mathbf{A}\): $$|\mathbf{r}(t)-\mathbf{L}| \ge |f(t)-L_1|, |\mathbf{r}(t)-\mathbf{L}| \ge |g(t)-L_2|, \text{ and } |\mathbf{r}(t)-\mathbf{L}| \ge |h(t)-L_3|$$ Since \(|\mathbf{r}(t)-\mathbf{L}| < \epsilon\) when \(0 < |t-a| < \delta\), it follows that \(|f(t)-L_1| < \epsilon, |g(t)-L_2| < \epsilon, \text{ and } |h(t)-L_3| < \epsilon\) when \(0 < |t-a| < \delta\). This implies that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text{ and } \lim _{t \rightarrow a} h(t)=L_{3}\). Part b:

Writing the definition of the limit for component functions

Given \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2}, \text{ and } \lim _{t \rightarrow a} h(t)=L_{3}\), this means that for any \(\epsilon_1, \epsilon_2, \epsilon_3 > 0\), there exists \(\delta_1, \delta_2, \delta_3 > 0\) such that \(|f(t)-L_1| < \epsilon_1\), \(|g(t)-L_2| < \epsilon_2\), and \(|h(t)-L_3| < \epsilon_3\) when \(0 < |t-a| < \delta_1, \delta_2, \delta_3\), respectively.

Define one epsilon for three components

Let \(\epsilon = \min\{\epsilon_1, \epsilon_2, \epsilon_3\}\) and \(\delta = \min \{\delta_1, \delta_2, \delta_3\}\). Then, we have \(|f(t)-L_1| < \epsilon, |g(t)-L_2| < \epsilon, \text{ and } |h(t)-L_3| < \epsilon\) when \(0 < |t-a| < \delta\).

Calculating the distance between r(t) and L

Now, we will calculate \(|\mathbf{r}(t)-\mathbf{L}|\). We can rewrite \(\mathbf{r}(t) - \mathbf{L} = \langle f(t)-L_1, g(t)-L_2, h(t)-L_3 \rangle\). Using the inequality \(||\mathbf{A}|| \le \sum |A_i|\), we get: $$|\mathbf{r}(t)-\mathbf{L}| \le |f(t)-L_1| + |g(t)-L_2| + |h(t)-L_3|$$ Since \(|f(t)-L_1| < \epsilon, |g(t)-L_2| < \epsilon, \text{ and } |h(t)-L_3| < \epsilon\), it follows that \(|\mathbf{r}(t)-\mathbf{L}| < 3\epsilon\) when \(0 < |t-a| < \delta\). Therefore, \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\).

Key concepts

Vector Functions

Vector functions are mathematical expressions that define a vector in terms of one or more variables. In essence, they allow us to describe quantities that have both magnitude and direction, using functions of a variable, typically denoted as \(t\). For example, in the exercise, we see \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\), representing a vector function composed of three component functions: \(f(t)\), \(g(t)\), and \(h(t)\).

Each component function describes a coordinate of the vector. This idea is prevalent in physics and engineering, where vector functions might represent positions, velocities, or other vector quantities.

One key aspect of vector functions is that they allow us to work with and analyze curves and surfaces in space. For instance, their derivatives can help determine tangents, velocities, and other dynamic properties.

Limits

Understanding limits is crucial in calculus, as they deal with the behavior of functions as they approach a particular point. In the context of vector functions, limits help us analyze how the components of the vector behave near a certain value of \(t\).

The exercise focuses on the statement \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}\), signifying that as \(t\) gets closer to \(a\), the vector function \(\mathbf{r}(t)\) approaches the vector \(\mathbf{L} = \langle L_1, L_2, L_3 \rangle\).

This concept can be decomposed into limits of individual components: if each component \(f(t), g(t), h(t)\) of a vector function has a limit at \(t = a\), the vector function itself also has a limit at that point. This approach simplifies complex vector problems into manageable scalar problems.

Componentwise Limit Theorem

The Componentwise Limit Theorem simplifies the evaluation of limits for vector functions by considering each component separately. This theorem states that the limit of a vector function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is \(\langle L_1, L_2, L_3 \rangle\) if and only if each component function approaches its corresponding limit: \(\lim _{t \rightarrow a} f(t) = L_1\), \(\lim _{t \rightarrow a} g(t) = L_2\), and \(\lim _{t \rightarrow a} h(t) = L_3\).

This theorem helps students by:

• Allowing one to split complex problems into simpler, single-variable calculus problems.
• Offering a clear pathway to solve vector function limits by leveraging known results for scalar functions.
• Highlighting the importance of each component in determining the overall behavior of vector functions.

Using this method makes the analysis of limits much more intuitive, ensuring that calculations with vector functions follow logically from familiar single-variable calculus results.

Inequalities in Calculus

Inequalities like \(|\mathbf{r}(t)-\mathbf{L}| < \epsilon\) play a significant role in proving limit statements in calculus. They provide a mathematical tool to express that the values of a function can be made arbitrarily close to a target value.

In the exercise, inequalities are used to illustrate how each component of the vector function \(\mathbf{r}(t)\) gets closer to its respective limit \(L_i\). By asserting \(|\mathbf{r}(t)-\mathbf{L}| < \epsilon\), we imply that the vector difference between \(\mathbf{r}(t)\) and \(\mathbf{L}\) can be smaller than any pre-selected small positive number, \(\epsilon\).

This method is doubly effective since:

• Establishes a strong foundation for the definition of limits.
• Ensures reliability in results by bounding function deviations from a limit.
• Uses familiar scalar inequalities to solve vector problems, enhancing understanding.

Working with inequalities sharpens understanding of limits and solidifies foundational calculus skills, easing the transition to more advanced topics.