### Video Transcript

Consider the function π of π₯ is
equal to one minus π₯ when π₯ is less than zero, zero when π₯ equals zero, and one
plus two π₯ when π₯ is greater than zero. Part one, what is π of zero?

For this question, weβve been given
a piece wise function defined by three different subfunctions. To begin, we simply must evaluate
π when π₯ equals zero. In fact, the second branch of our
subfunction defines this telling us that π is zero when π₯ equals zero. We can, therefore, simply state
that πof zero is equal to zero. And we have answered the first part
of our question.

Part two, what is the limit as π₯
approaches zero from the left of π of π₯?

Now for this part of the question,
since π₯ is approaching zero from the left, we know that π₯ is less than zero. And hence, π of π₯ is defined by
our first subfunction, one minus π₯. When finding our limit, we can,
therefore, replace π of π₯ with their subfunction. And we can then solve this by
taking a direct substitution approach to get one minus zero, which is of course
equal to one.

Now, here we might notice that the
third part of our question is very similar, instead asking us for the right-sided
limit as π₯ approaches zero of π of π₯.

When π₯ is greater than zero, π of
π₯ is defined by the third subfunction, one plus two π₯. We can evaluate this one-sided
limit in the same way, in putting our subfunction as π of π₯ and, again, taking a
direct substitution approach of π₯ equals zero to find that our limit is equal to
one. We have now answered part two and
three of the question, finding that both of our one-sided limits as π₯ approach zero
are equal to one.

Finally, part four of the question
asks, what type of discontinuity does the function π have at π₯ equals zero?

For this part of the question, we
first restate that both of the one-sided limits that we found exist are finite and
are equal to each other. Putting these two bits of
information together, we are also able to conclude that the normal limit as π₯
approaches zero also exists and is finite taking the same value, one. Letβs now look back at our answer
to part one. We found that when π₯ equals zero,
π is also zero. In other words, π of zero equals
zero. Again, letβs combine our
information. We have found that the limit as π₯
approaches zero of π of π₯ exists and is finite but is not equal to π of zero.

We now recall that this is the
exact condition that must be satisfied for a removal discontinuity at π₯ equals
zero. With this, we have answered all
four parts of our question. When π₯ equals zero, we have
evaluated our function, found its limits, and classified the type of discontinuity
that occurs. As a final note, if we were to
graph our function, it might look a little bit like this. And we could get rid of our removal
discontinuity by redefining π of zero equals one.