Limit of Vector Valued Function

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

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

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

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

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

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

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

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

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.

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 \(t=4.\)

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.