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Classifying Discontinuities
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In this lesson, weβll learn how to identify the different types of function discontinuity at a given point.
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When thinking about discontinuities, itβs first helpful to recap the condition for continuity at a point.
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This is that the limit as π₯ approaches π of some function π of π₯ is equal to the same function, evaluated when π₯ equals π, π of π.
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Of course, the implications of this are that the left and the right limits as π₯ approaches π of π of π₯ must exist and agree and that π must be in the domain of the function.
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So π of π must be defined.
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When the function π of π₯ does not satisfy the continuity condition, we say that our function is not continuous.
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If this happens at a point, we say that we have a discontinuity.
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Let us now examine the various different ways in which this can happen.
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Our first case is that of a removable discontinuity.
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This occurs when the limit as π₯ approaches π of π of π₯ exists and is finite.
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However, π of π is not equal to the value of this limit.
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To illustrate this type of discontinuity, here is the example function π one of π₯.
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We can clearly see that π₯ equals three is not in the domain of our function.
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And this is shown by the hollow dot on our graph at this point.
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π one of three is not defined.
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And hence this is our removal discontinuity.
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Before proceeding, let us look at another example, function π two, which is defined in the same way as π one when π₯ is not equal to three but is defined as one when π₯ is equal to three.
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Now for both π one and π two, the limit as π₯ approaches three is equal to four.
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In the case of π one, our function was undefined when π₯ equals three.
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In the case of π two, our function is equal to one when π₯ equals three.
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This is denoted by the solid dot on our graph.
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In both cases, however, this is not equal to the value of the limit.
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This illustrates that our function may or may not be defined at the point where our removable discontinuity exists.
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A final tip which might help you remember, these discontinuities are named as such because we could remove the discontinuity by redefining the function at a single point, in both of these cases, the point where π₯ is equal to three.
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The next type of discontinuity weβll look at is called an essential discontinuity.
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And this is sometimes called a nonremovable discontinuity.
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These occur when either the left, the right, or both of the one-sided limits as π₯ approaches π of π of π₯ do not exist.
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Here, it is worth remembering that when we say that a limit is equal to a positive or a negative infinity, this is simply a particular way of expressing that the limit does not exist.
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Expressing the limit in this way does, however, give us useful information about our function.
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So weβll add it as a side note to our definition.
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To illustrate, letβs look at an example function π three, which is one over π₯.
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As π₯ approaches zero from the left, our function will approach a negative infinity.
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And as π₯ approaches zero from the right, our function will approach positive infinity.
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Again, since infinity is a concept and not a number, these limits do not exist.
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This satisfies our criteria and we therefore have an essential discontinuity at π₯ equals zero.
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Itβs worth noting that even if we used a different example function, π four, in which the left- and the right-sided limits approach the same infinity, which in this case is positive infinity, and we were able to say that the normal limit as π₯ approaches zero is equal to infinity.
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This still satisfies that criteria for an essential discontinuity at π₯ equals zero, since none of these limits exist.
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Let us now look at a different example function.
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In the case of a function such as sin of one over π₯ as π₯ approaches zero from either the left or the right, the value of the function itself will oscillate more and more rapidly between negative one and one.
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Given this information, it doesnβt make sense for us to assign a value to either the left- or the right-sided limits, since π five of π₯ seems to be approaching two values simultaneously.
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Since we say that these limits do not exist, this again satisfies that criteria for an essential discontinuity at π₯ equals zero.
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As a final note, although both of these cases are indeed essential discontinuities, we sometimes use more specific language when referring to them, calling the first case an infinite discontinuity and the second case an oscillating discontinuity.
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The final case weβll look at is called a jump discontinuity.
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This occurs when both of the one-sided limits as π₯ approaches π of π of π₯ exist and are finite but are not equal to each other.
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You will often, but not always, see a jump discontinuity when a function defined piecewise, as with the example function π six shown here.
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If we examine the boundary between our two subfunctions, which occurs when π₯ equals two, we can clearly see that as we approach from the left, π of π₯ approaches three and as we approach from the right, π of π₯ approaches two.
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We might also note the position of the hollow dot and the filled dot which tells us that π of two is defined here and is equal to two.
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But actually for a jump discontinuity, weβre not particularly interested in this fact.
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Since both of our one-sided limits exist and are finite but are not equal to each other, weβve already satisfied the condition for a jump discontinuity.
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In fact, this would be the same case, even if π of two were undefined, the more important feature being the jump that we observe at π₯ equals two.
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Okay, now that weβve seen the different types of discontinuity, let us look at an example.
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Find the type of discontinuity that the function π has at π₯ equals zero, π₯ equals two, π₯ equals five, and π₯ equal six if it has any discontinuity at these points.
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For this question, weβve been given a graph and asked to identify whether the discontinuities exist at certain points and to classify them if so.
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To answer this question, let us go through the different types of discontinuity that weβre aware of, specifically the features of each that we would observe on a graph.
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The first type of discontinuity that we know is a removable one.
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This is when the limit as π₯ approaches π of π of π₯ exists and is finite, equaling some value πΏ here.
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But π of π is not equal to this value, πΏ.
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Looking at our graph, we see that as π₯ approaches two from the left and from the right, π of π₯ approaches one.
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In other words, the limit as π₯ approaches two of π of π₯ is equal to one.
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Another thing we notice when π₯ equals two is that π of two is not defined to be one, denoted by the hollow dot, but rather is defined to be negative one, denoted by the solid dot.
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In other words, π of two is equal to negative one.
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Weβve now found that the limit as π₯ approaches two of π of π₯ is not equal to π of two.
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And this is the condition for a removable discontinuity.
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We have, therefore, answered part two of the question.
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Let us now move on to the next type of discontinuity, an essential this continuity.
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This occurs when either the left, the right, or both of the one-sided limits as π₯ approaches π of π of π₯ do not exist.
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Looking at the graph, we see that as π₯ approaches six from both the left and the right, we see that the value of π appears to approach negative infinity.
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Here we recall that when we say that a limit is equal to infinity, whether positive or negative, this is just a particular way of expressing that the limit does not exist, since infinity is a concept instead of a number.
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Given this information, we have satisfied the condition for an essential discontinuity, which is that at least one of our one-sided limits must not exist.
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And in this case, in fact, both do not.
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We have therefore found that, at π₯ equal six, we have an essential discontinuity.
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Finally, let us think about jump discontinuities.
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These occur when both of the one-sided limits as π₯ approaches π of π of π₯ exist and are finite but are not equal to each other.
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Again, we look at our graph to find the cases where this might be true.
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Observing π₯ equal zero, we see that as π₯ approaches from the left, the value of π also approaches zero, whereas as π₯ approaches zero from the right, the value of π approaches three.
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Both of our one-sided limits as π₯ approaches zero of π of π₯ do exist and are finite.
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However, they are not equal to each other, which is the condition for a jump discontinuity.
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We, therefore, found that π has a jump discontinuity when π₯ equals zero.
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To finish off this question, we must evaluate the point where π₯ is equal to five.
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As π₯ approaches five from both the left and the right, the value of π approaches three.
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Since both of the one-sided limits both exist and agree, we can also say the normal limit exists and takes this same value.
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Another thing we can see is that π of five is defined to be three, as seen by the solid dot on our graph.
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Now, with these two bits of information, we have found that the limit as π₯ approaches five of π of π₯ is equal to π of five.
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You may recognize this as the continuity condition.
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And weβve, therefore, proved that π is continuous when π₯ is equal to five.
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We do have a sharp corner at this point, which means that π is not differentiable.
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However, this is outside the scope of this video.
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Since weβve proved continuity here, by definition, we can say that π has no discontinuity when π₯ equals five.
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With this information, weβve completed the question and weβve identified all the discontinuities shown on the graph of π.
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Okay, weβve seen some graphical examples of discontinuities.
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But now letβs consider algebraic examples where no visual cues are given.
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The first thing we should be aware of is when dealing with a piecewise function, it is always worthwhile to check and evaluate the boundaries between the different subfunctions.
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Letβs have a look at an example of this now.
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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.
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Part one, what is π of zero?
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For this question, weβve been given a piece wise function defined by three different subfunctions.
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To begin, we simply must evaluate π when π₯ equals zero.
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In fact, the second branch of our subfunction defines this telling us that π is zero when π₯ equals zero.
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We can, therefore, simply state that πof zero is equal to zero.
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And we have answered the first part of our question.
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Part two, what is the limit as π₯ approaches zero from the left of π of π₯?
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Now for this part of the question, since π₯ is approaching zero from the left, we know that π₯ is less than zero.
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And hence, π of π₯ is defined by our first subfunction, one minus π₯.
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When finding our limit, we can, therefore, replace π of π₯ with their subfunction.
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And we can then solve this by taking a direct substitution approach to get one minus zero, which is of course equal to one.
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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₯.
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When π₯ is greater than zero, π of π₯ is defined by the third subfunction, one plus two π₯.
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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.
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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.
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Finally, part four of the question asks, what type of discontinuity does the function π have at π₯ equals zero?
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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.
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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.
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Letβs now look back at our answer to part one.
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We found that when π₯ equals zero, π is also zero.
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In other words, π of zero equals zero.
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Again, letβs combine our information.
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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.
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We now recall that this is the exact condition that must be satisfied for a removal discontinuity at π₯ equals zero.
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With this, we have answered all four parts of our question.
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When π₯ equals zero, we have evaluated our function, found its limits, and classified the type of discontinuity that occurs.
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As a final note, if we were to graph our function, it might look a little bit like this.
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And we could get rid of our removal discontinuity by redefining π of zero equals one.
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Okay, weβve just seen a piecewise function, but in many cases our function will not be defined this way.
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Another thing we should be on the lookout for is rational functions or functions with quotients of the form π of π₯ is equal to π of π₯ over π of π₯.
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In particular, we should pay attention to values of π₯, which will make our denominator here π of π₯ equal to zero.
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At these values of π₯, we would be dividing by zero.
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And hence, we know that our function π of π₯ would be undefined.
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Itβs worth noting that when we see that a denominator of zero is possible at some value of π₯, we cannot immediately conclude the type of discontinuity that will exist here.
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We can look back at two of the examples that we saw earlier to illustrate this.
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For π one, the denominator of the quotient will be zero when π₯ equals three.
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For π four, the denominator of the quotient will be zero when π₯ equals zero.
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Although in both cases we observe a denominator of zero, the first is a removal discontinuity and the second is an essential, an infinite discontinuity.
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In order to algebraically distinguish between the two different types, we should still be examining our limits as outlined by the criteria earlier.
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Now another point, in both of these cases, it is very easy to find the value of π₯ which will make our denominator zero.
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In some other cases, however, it might not be immediately obvious.
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Weβre still considering π of π₯ is equal to π of π₯ over π of π₯.
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But here, letβs imagine that π of π₯ is some polynomial.
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One tool that we can use to help find that π of π₯ equals zero is the factor theorem.
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This tells us that if π₯ minus π is a factor of π of π₯, then π of π must be zero.
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As a quick example, if π of π₯ was some quadratic function and we were able to factorise it as π₯ minus π multiplied by π₯ minus π, this means that the denominator of our function π of π₯ would be zero when π₯ equals π and π₯ equals π.
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And hence, we should be evaluating our function π for discontinuities at these values of π₯.
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Letβs see an example.
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Consider the function π of π₯ is equal to π₯ minus five divided by π₯ squared minus three π₯ minus 10.
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Find all the values of π₯ at which π has discontinuities.
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Find the type of each discontinuity.
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To begin this question, we note that π of π₯ is a rational function in the form π of π₯ over π of π₯.
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When seeing this form, we might recall that our next step is to look for values of π₯, which would make our denominator here called π of π₯ equal to zero.
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One tool that we can use to help us find these values is the factor theorem.
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This tells us that if π₯ minus π is a factor of π of π₯, then π of π must be equal to zero.
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Our first step should then be to try and factorise our denominator.
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With a bit of inspection, we can factorise our denominator to π₯ plus two times π₯ minus five.
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Here, the factor theorem tells us that when π₯ equals negative two or when π₯ equals five, our denominator would be zero.
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And hence π of π₯ would be undefined.
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We can extend our reasoning a little further to say that these are the values at which weβll find our discontinuities.
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But weβll need to look a little closer to find their type.
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Letβs begin with π₯ equals negative two.
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If we were to take the limit as π₯ approaches negative two π of π₯, using the factorised form of our denominator, and we were to take a direct substitution approach, of course, we would find that our limit does not exist, since we already know that we have a divide by zero.
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Again here, finding that a limit is equal to a positive or negative infinity is a particular way of expressing that limit doesnβt exist.
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From this information alone, we could conclude that either the left, the right, or both of the one-sided limits also do not exist.
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However, letβs take a closer look at them to see whatβs going on.
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We take a similar approach here, but we noticed that since both of these parentheses are equal and nonzero, we can cancel them out.
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We also noticed that, in the denominator of our quotient, we have a negative two which weβre approaching from the left and add two.
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This means that we have a one divided by zero.
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But weβre approaching the zero from the left.
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Since weβre approaching zero from the left, we have a negative number.
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And if we divide one by an infinitesimally small negative number, we get a negative infinity.
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This is a loose way in which we can understand whether our infinity is a positive or a negative.
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We could follow this same process to find that our right-sided limit would be a positive infinity.
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Although these last two steps were not strictly necessary, it does give us a better understanding as to what our function will do as π₯ approaches negative two from the left and from the right, respectively.
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Here, we have found that both our left- and our right-sided limit as π₯ approaches negative two does not exist.
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And hence, at this point, we have an essential discontinuity.
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And given our working, we could more specifically classify this as an infinite discontinuity.
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Let us now move on to the case where π₯ equals five.
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We evaluate the limit as π₯ approaches five of π of π₯.
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This time, direct substitution leads us to the indeterminate form of zero over zero.
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And weβll be using a different trick to get around this.
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So far, weβve almost been ignoring the fact that our original function π of π₯ appears to have a common factor of π₯ minus five on the top and bottom half of the quotient.
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If we were to cancel this factor, we would be left with one over π₯ plus two.
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Now here, we must be very careful not to say this is π of π₯.
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And instead, weβll call it something different, say π of π₯.
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Although it seems like π of π₯ and π of π₯ are equal, this is only true when π₯ is not equal to five, since we know that π₯ equals five is not in the domain of π of π₯.
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Yet it is in the domain of π of π₯.
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Herein, lies our trick.
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Since the limit concerns values of π₯ which are arbitrarily close to five but not where π₯ is actually equal to five, weβre able to say that the limit as π₯ approaches five of π of π₯ is equal to the limit as π₯ approaches five π of π₯.
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This now allows us to use a direct substitution approach since π₯ equals five is in the domain of π of π₯.
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Doing so, we find that the value of our limit is one over seven.
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Weβre now in a position to think about our discontinuity.
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We found that the limit as π₯ approaches five of π of π₯ exists and is finite, taking the value of one over seven.
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However, we concluded earlier that when π₯ equals negative two or when π₯ equals five, π of π₯ was undefined since we had a divide by zero.
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Hence, π of five is undefined.
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Putting these two bits of information together, we have found that the limit as π₯ approaches five of π of π₯ exists, is finite, but is not equal to π of five.
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This is the exact condition for a removable discontinuity at π₯ equals five.
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We have now fully answered our question.
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We have found that π has an essential and infinite discontinuity at π₯ equals negative two.
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And π has a removal discontinuity at π₯ equals five.
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If we were to sketch the graph of π of π₯, it might look a little like this.
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To finish off our video.
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Letβs go through some key points.
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When a function does not satisfy the continuity condition, it is said to be discontinuous.
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For discontinuities occurring at a point, we have learned to classify them as the following, firstly, removable discontinuities; secondly, essential discontinuities, which confer to be subcategorized into infinite and oscillating discontinuities; and finally, jump discontinuities.
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Each of the different types of discontinuity has their own set of conditions outlined here.
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When looking for discontinuities, it is useful to evaluate certain points of a function.
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For piecewise defined functions, look at values of π₯ at the boundary between the different subfunctions.
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And for quotients such as rational functions in the form of π of π₯ over π of π₯, look for values of π₯ where the denominator, π of π₯, equals zero.
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As a final note, in some cases where a function is defined using absolute values or noninteger powers of π₯, sometimes reexpressing this function into a more manageable form, which could perhaps be piecewise, might make discontinuities easier to find and to classify.