In this section we need to talk briefly about limits,
derivatives and integrals of vector functions.
As you will see, these behave in a fairly predictable manner. We will be doing all of the work in but we can naturally extend the formulas/work
in this section to (i.e.
ndimensional space).
Let’s start with limits.
Here is the limit of a vector function.
So, all that we do is take the limit of each of the
component's functions and leave it as a vector.
Example 1 Compute
where .
Solution
There really isn’t all that much to do here.
Notice that we had to use L’Hospital’s Rule on the y component.

Now let’s take care of derivatives and after seeing how
limits work it shouldn’t be too surprising that we have the following for
derivatives.
Example 2 Compute
for .
Solution
There really isn’t too much to this problem other than
taking the derivatives.

Most of the basic facts that we know about derivatives still
hold however, just to make it clear here are some facts about derivatives of
vector functions.
Facts
There is also one quick definition that we should get out of
the way so that we can use it when we need to.
A smooth curve is
any curve for which is continuous and for any t
except possibly at the endpoints. A
helix is a smooth curve, for example.
Finally, we need to discuss integrals of vector
functions. Using both limits and
derivatives as a guide it shouldn’t be too surprising that we also have the
following for integration for indefinite integrals
and the following for definite integrals.
With the indefinite integrals we put in a constant of
integration to make sure that it was clear that the constant in this case needs
to be a vector instead of a regular constant.
Also, for the definite integrals we will sometimes write it
as follows,
In other words, we will do the indefinite integral and then
do the evaluation of the vector as a whole instead of on a component by
component basis.
Example 3 Compute
for .
Solution
All we need to do is integrate each of the components and
be done with it.

Example 4 Compute
for .
Solution
In this case all that we need to do is reuse the result
from the previous example and then do the evaluation.
