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Limit of Vector Valued Function

- Calculus
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Suppose you throw a ball through the air. To get from the beginning of the path to the end, the ball must travel through every point in the path, it cannot teleport from one part of the path to another. This is because the path of a ball is **continuous**. In this example, since the ball is moving in 3D space, it can be represented as a 3-dimensional vector, and its path as a vector-valued function.

The limit of a vector-valued function can be defined in a very similar way to the limit of a scalar function.

Let \(I\) be an open interval with a point \(c \in I\) and a vector-valued function \(\vec{r}(t)\) which is defined at every point in \(I\) except possibly \(c.\) The** limit** of \(\vec{r}(t)\) as \(t\) approaches \(c\) is a vector

\[ \lim\limits_{t \rightarrow c} \vec{r}(t) = \vec{L} \]

such that for any \( \epsilon > 0, \) there exists a \( \delta > 0 \) such that for any \( t \neq c, \) if \( | t - c | < \delta ,\) then \( \| \vec{r}(t) - \vec{L} \| < \epsilon. \)

It is important to note that in this equation \(| t - c|\) will be the standard absolute value of a real number, while \(\| \vec{r}(t) - \vec{L} \|\) will be the modulus of a vector. This is very similar to the definition of a limit for a 1D function, the key difference being that for a 1D function \(f(x),\) \( \| \vec{r}(t) - \vec{L}\|\) will be replaced with the absolute value \( | f(x) - L|. \)

In the diagram above, you can see that whenever \(t\) is between \(c - \delta\) and \(c + \delta,\) \(\vec{r}(t)\) is within the circle with radius \(\epsilon\) of the point at which \(t=c.\) If this is true for every possible \(\epsilon\) you could choose, then the limit must exist.

Finding a limit just using its definition can be very tedious, so the following theorem makes it much simpler.

Given a vector-valued function \( \vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} \),

\[\lim\limits_{t \rightarrow c} \vec{r}(t) \]

exists if and only if

\[\lim\limits_{t \rightarrow c} f(t) \text{ and } \lim\limits_{t \rightarrow c} g(t) \]

exist. If

\[ \lim\limits_{t \rightarrow c} \vec{r}(t) \]

exists, then

\[ \lim\limits_{t \rightarrow c} \vec{r}(t) = \lim\limits_{t \rightarrow c} f(t) \vec{i} + \lim\limits_{t \rightarrow c} g(t) \vec{j}. \]

This is true for any finite-dimensional vector.

From this, finding the limit of an \(n\) dimensional vector-valued function just becomes finding the limit of \(n\) real-valued functions. To see how to find the limit of a real-valued function, see Limits. This rule makes sense visually when written in component form as above, as it just looks like the standard sum rule of limits. Let's look at an example of finding the limit of a vector-valued function in this way.

Deduce whether the following vector-valued function has a limit as \(t \) goes to \(0,\) and if so, the value of this limit.

\[ \vec{r}(t) = \begin{bmatrix} t^2 + 4t + 3 \\ e^t \end{bmatrix} \]

**Solution**

Let's first start by finding the limit of the first function, \(f(t) = t^2 + 4t + 3. \) The limit of a polynomial will always exist, thus this limit must exist. The limit as \(t\) goes to \(0\) of this function will be \(f(t) = 3.\)

The second function, \(g(t) = e^t,\) is a function that you may have seen the limit of before. The limit of this function as \(t\) goes to \(0\) is \(1\).

Since both limits exist, the limit of the vector-valued function must also exist, and it is

\[ \lim\limits_{t \rightarrow 0} \vec{r}(t) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. \]

Just as there are rules of the limits of real-valued functions, there are also rules to make finding the limits of vector-valued functions easier too. For any scalars \(a\) and vector-valued functions \(\vec{r}_1(t), \vec{r}_2(t)\) defined on an open interval \(I\) which contains the point \(c,\) the following rules apply:

**Sum rule:**\[ \lim\limits_{t \rightarrow c} (\vec{r}_1(t) + \vec{r}_2(t)) = \lim\limits_{t \rightarrow c} \vec{r}_1(t) + \lim\limits_{t \rightarrow c} \vec{r}_2(t). \]**Scalar multiple rule:**\[ \lim\limits_{t \rightarrow c} (a \vec{r}_1(t)) = a \lim\limits_{t \rightarrow c} (\vec{r}_1(t)). \]**Limit of dot product:**\[ \lim\limits_{t \rightarrow c} (\vec{r}_1(t) \cdot \vec{r}_2(t)) = (\lim\limits_{t \rightarrow c} \vec{r}_1(t)) \cdot ( \lim\limits_{t \rightarrow c} \vec{r}_2(t) ). \]**Limit of scalar product:**\[ \lim\limits_{t \rightarrow c} (\vec{r}_1(t) \times \vec{r}_2(t)) = (\lim\limits_{t \rightarrow c} \vec{r}_1(t)) \times ( \lim\limits_{t \rightarrow c} \vec{r}_2(t) ). \]

The sum rule and scalar multiple rule here are exactly the same as the sum rule and scalar multiple rule of scalar-valued functions, that you will have seen when you first encountered limits. The product rule of limits for scalar-valued functions appears here twice as the limit of dot product rule and limit of scalar product rule, since in vector spaces we have two methods of multiplication.

You may notice that the quotient rule of limits for scalar-valued functions has no vector-valued counterpart, and this makes sense because two vectors cannot be divided by each other. Similarly, no equivalent of L'Hopital's rule exists for vector-valued functions. It is important to remember that when finding the limits of the individual component functions, you can use the quotient rule and L'Hopital's rule, since then you are working with scalar-valued functions.

Remember the definition of continuity for a scalar function: \(\vec{r}(t)\) is continuous at \(c\) if

\[\lim\limits_{t \rightarrow c} \vec{r}(t) = \vec{r}(c).\]

Just as the definition of continuity of a real-valued function is defined by the limit of the function being equal to the value of the function at that point, the continuity of a vector-valued function is defined in the same way.

Let \(I\) be an open interval with a point \(c \in I\) and a vector-valued function \(\vec{r}(t)\) which is defined on \(I.\)

Then \(\vec{r}(t)\) is **continuous at \(c\)** if

\[\lim\limits_{t \rightarrow c} \vec{r}(t) = \vec{r}(c),\]

and \(\vec{r}(t)\) is **continuous on \(I\) **if at every point \(t' \in I,\)

\[ \lim\limits_{t \rightarrow t'} \vec{r}(t) = \vec{r}(t').\]

This definition of continuity is exactly the same as the normal definition of continuity for a real-valued function.

The graph above shows a function that is discontinuous at \(t = c\). This function does not have a defined value here, and hence is cannot be continuous.

Let's look at some examples of finding the limits of vector-valued functions.

Deduce whether the following vector-valued function has a limit as \(t \) goes to \(0,\) and if so, the value of this limit.

\[ \vec{r}(t) = \begin{bmatrix} \frac{\sin{t}}{t} \\ 4 \end{bmatrix} \]

**Solution:**

Let's first start by finding the limit of the first function,

\[f(t) = \frac{\sin{t}}{t}. \]

There are many ways to prove this, but let's use L'Hopital's Rule. Since

\[ \lim\limits_{t \rightarrow 0} \sin{t} = \lim\limits_{t \rightarrow 0} t = 0, \]

LHopital's rule can be used. This means that:

\[ \begin{align} \lim\limits_{t \rightarrow 0} \frac{\sin(t)}{t} & = \lim\limits_{t \rightarrow 0} \frac{(\sin(t))'}{(t)'} \\ & = \lim\limits_{t \rightarrow 0} \frac{\cos(t)}{1} \\ &= 1. \end{align} \]

Next, find the limit of the second term. Since this is just a constant, \(4\), the limit as \(t\) goes to \(0\) will also be \(4\).

Finally, calculate the limit of the final term,

\[ \frac{t^2 -4t}{t}. \]

By taking out a factor of \(t\) from the top fraction, you can cancel it out and simplify this fraction.

\[ \frac{t^2 - 4 t}{t} = \frac{t(t - 4)}{t} = t - 4. \]

Now, you can take the limit of this as \(t\) goes to \(0\), and see that this will become \(-4.\) Since all three limits exist, the limit of the vector-valued function must also exist, and it has the value

\[ \lim\limits_{t \rightarrow 0} \vec{r}(t) = \begin{bmatrix} 1 \\ 4 \end{bmatrix}. \]

Let's look at another example.

Deduce whether the following vector-valued function has a limit as \(t \) goes to \(0,\) and if so, the value of this limit.

\[ \vec{r}(t) = \begin{bmatrix} \frac{t^2 - 4t}{t} \\ t+4 \end{bmatrix}. \]

**Solution:**

First, find the limit of the first component function. This is \( \frac{t^2 -4t}{t}. \) By taking out a factor of \(t\) from the top fraction, you can cancel it out and simplify this fraction. This can be done because you are limiting \(t\) to \(0,\) and thus \(t\) is never actually \(0\). So

\[ \begin{align} \frac{t^2 - 4 t}{t} &= \frac{t(t - 4)}{t} \\ &= t - 4. \end{align}\]

Now, you can take the limit of this as \(t\) goes to 0, and see that it is \(-4.\)

Next, find the limit of the second function, which is \(t+4.\) This is a standard linear function, and hence you can find the limit by just plugging in the value, Hence, the limit is \(4.\) Since both limits exist, the limit of the vector-valued function must exist, and this limit must be

\[ \lim\limits_{t \rightarrow 0} \vec{r}(t) = \begin{bmatrix} -4 \\ 4 \end{bmatrix}. \]

Let's look at one final example of limits of vector-valued functions.

\[ \vec{r}(t) = \sin{\frac{1}{t}} \vec{i} + \frac{t^2 + 4t + 8}{t+4} \vec{j} \]

**Solution:**

Find the individual limits of the functions in each component, and see if they exist. The first function is

\[ \sin{\frac{1}{t}}. \]

As \(t\) goes to 0, \(\frac{1}{t}\) will increase at a faster rate, making the sine function oscillate between \(-1\) and \(1.\) Given any \(\epsilon\) between \(0\) and \(1,\) there is no \(\delta\) such that \( \| \vec{r}(t) - \vec{L} \| < \epsilon \) is true if \( | t - c | < \delta. \) Since this limit does not exist, the limit of the whole vector-valued function must not exist.

Let's look at some examples where you must deduce whether a function is continuous or not.

Deduce whether or not the following vector-valued function is continuous at the point \(t = 4:\)

\[ \vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} \]

where

\[ \begin{align} f(t) & = 8t \\ g(t) & = \begin{cases} \frac{t^2-16}{t-4} &\quad t \neq 4 \\ 8 &\quad t = 4 \end{cases} \end{align} \]

**Solution:**

First, find \(\vec{r}(4).\) Firstly \(f(4)\) will be \(8 \cdot 4 = 32.\) Since \(t=4,\) \(g(t) = 8\) by it's definition. Hence,

\[ \vec{r}(4) = 32 \vec{i} + 8 \vec{j}. \]

Now, find \(\lim\limits_{t \rightarrow 4} \vec{r}(t).\) You can do this by finding the individual limits again. The limit of \(f(t)\) is simply just the value of \(f(t),\) since this is a linear equation. Hence,

\[\lim\limits_{t \rightarrow 4} f(t) = 32.\]

The limit

\[\lim\limits_{t \rightarrow 4} g(t)\]

is a little more complicated. Since the limit will never allow \(t\) to actually take on the value of \(4,\) it must be that

\[g(t) = \frac{t^2 - 16}{t-4}. \]

You can factor the top polynomial to become

\[ g(t) = \frac{(t+4)(t-4)}{t-4}, \]

and since \(t\) is not \(4,\) \(t-4 \neq 0.\) This means you can cancel out the \(t-4\) on the top and bottom.

\[ g(t) = t+4\]

Now, since this is a linear function, you can take the limit by simply substituting in the value of \(4\) to get

\[ \lim\limits_{t \rightarrow 4} g(t) = 8. \]

Thus, the limit of the function is:

\[ \lim\limits_{t \rightarrow 4} \vec{r}(t) = 32 \vec{i} + 8 \vec{j} = \vec{r}(4). \]

Since these are equal, the function must be continuous at \(t=4.\)

Let's look at another example, this time where you must find continuity of the whole vector-valued function.

Prove that the following vector-valued function \( \vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} \) is continuous on the domain \([-1,1],\) where:

\[ \begin{align} f(t) & = t^2 + 6t + 2 \\ g(t) & = \frac{\tan{t}}{t}. \end{align} \]

**Solution:**

First, see if the function components of the vector are continuous. The first function is \(f(t) = t^2 + 6t + 2.\) Polynomials are always continuous on any domain, and hence they will be continuous on the given domain \((-1,1).\)

Second is the function \(g(t) = \frac{\tan{t}}{t}.\) This is continuous everywhere on the domain with the exception of the point \(t = 0.\) The limit at this point is \(1\), so it could be made to be continuous by modifying the function to be:

\[ g'(t) \begin{cases} \frac{\tan{t}}{t} &\quad t \neq 0 \\ 1 &\quad t = 0 \end{cases}, \]

however, since this is a different function, \(g(t)\) is not continuous. Since \(g(t)\) is not continuous, the whole vector-valued function is not continuous.

- The limit of \(\vec{r}(t)\) as \(t\) approaches \(c\) is a vector \( \lim\limits_{t \rightarrow c} \vec{r}(t) = \vec{L} \) such that for any \( \epsilon > 0, \) there exists a \( \delta > 0 \) such that for any \( t \neq c, \) if \( | t - c | < \delta ,\) then \( \| \vec{r}(t) - \vec{L} \| < \epsilon. \)
- The limit of a vector-valued function \(\vec{r}(t) = f(t) \vec{i} + g(t) \vec{j} + h(t) \vec{k}\) exists if and only if the limits of \(f(t), g(t) \) and \(h(t)\) exist. If it does exist then, \[ \lim\limits_{t \rightarrow c} \vec{r}(t) = \lim\limits_{t \rightarrow c} f(t) \vec{i} + \lim\limits_{t \rightarrow c} g(t) \vec{j} + \lim\limits_{t \rightarrow c} h(t) \vec{k}. \]
- \(\vec{r}(t)\) is
**continuous at \(c\)**if \(\lim\limits_{t \rightarrow c} \vec{r}(t) = \vec{r}(c).\) - \(\vec{r}(t)\) is
**continuous**if at every point \(t' \in I,\) \( \lim\limits_{t \rightarrow t'} \vec{r}(t) = \vec{r}(t').\)

The limit of a vector-valued function **r**(t) as t goes to c is a value L such that

for all ε > 0, there exists a **𝛿** > 0 such that for all t not equal to c, if | t - c | < **𝛿,** then | r(t) - L | < ε.

A vector-valued function **r**(t) is continuous at a point c if

lim_{t→ c }**r**(t) = **r**(c).

**r**(t) is continuous if the above statement is true for every c in the domain of **r**.

The limit of a vector-valued function

**r**(t) = f(t) **i **+ g(t) **j** + h(t) **k** will be

lim **r**(t) = lim f(t) **i **+ lim g(t) **j** + lim h(t) **k**

A vector-valued function is continuous at a point c if

lim_{t→ c} **r**(t) = **r**(c).

The limit as t goes to 0 of the vector-valued function

r(t) = (sin(t))/(t) **i** + 4 **j** + t **k**

is 1 **i** + 4 **j** + 0**k**. The function, however, is not continuous at t = 0 because the **i** term is not defined at t=0, since it would require dividing by 0.

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