# Question Video: Evaluating the Type of Discontinuity of a Function at a Point Mathematics

Consider the function π(π₯) = 1 β π₯ when π₯ < 0, π(π₯) = 0 when π₯ = 0 and π(π₯) = 1 + 2π₯ when π₯ > 0. (i) What is π(0)? (ii) What is lim_(π₯ β 0β») π(π₯)? (iii) What is lim_(π₯ β 0βΊ) π(π₯)? (iv) What type of discontinuity does the function π have at π₯ = 0?

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### 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, weβve 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 πof 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. Letβs 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, letβs 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.