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In this video, weβre going to learn about continuity at a point.
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This is a step on the way to understanding continuous functions, functions like polynomial functions, exponential functions, and certain trigonometric functions.
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Whose graphs can be drawn in a single stroke of the pen.
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You may have noticed that, for a lot of functions π, the limit of π of π₯ as π₯ approaches some value π is just π evaluated at π, π of π.
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A function π is said to be continuous at the point π if this happens.
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If the limit of π of π₯ as π₯ approaches π is just π of π.
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This is the definition of continuity at a point.
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We can see then that the function π is continuous at the point π if direct substitution works to find the limit of π of π₯ as π₯ approaches π.
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To understand continuity at a point, we need to understand how a function can fail to be continuous at the point π.
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How this equation can fail to hold true.
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There are several things that can go wrong.
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For example, this limit on the left-hand side could not exist.
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For π to be continuous at π then, we need this limit to exist.
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Similarly, we need the right-hand side of the equation to exist.
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π must be defined at π.
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If it isnβt, then the right-hand side of our equation, thatβs π of π, is undefined.
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And so our equation canβt possibly hold.
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Another way of saying that π must be defined at π is to say that π must be in the domain of π.
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What else could go wrong?
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Well, the limit on the left-hand side could exist.
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And the function π of π could be defined.
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But these values could be different.
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We need the values of the limit of π of π₯ as π₯ approaches π and π of π to be equal.
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These are the three things we needed to check to show that a function π is continuous at a number π.
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Letβs now apply this checklist to some examples.
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For convenience, weβll always check that π is defined at π before checking that the limit as π of π₯ approaches π exists.
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And so the checklist for our examples has a slightly different order, which Iβd encourage you to use too.
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Given π of π₯ equals π₯ squared plus π₯ minus two all over π₯ minus one, if possible or necessary, define π of one so that π is continuous at π₯ equals one.
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So we have a function π, which is a rational function.
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And weβd like π to be continuous at the point π₯ equals one.
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And weβre told that we should define π of one to make this so, but only if it is possible or necessary.
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What does that mean?
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Well, if π is already continuous at the point π₯ equals one, then it isnβt necessary to define π of one to make this so.
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Itβs already been done for us.
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On the other hand, π could be discontinuous at the point π₯ equals one.
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But in the way that itβs not possible to things just by defining π of one.
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There could be some bigger issue stopping π from being continuous at this point.
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Letβs first check whether it is necessary to define π of one to make π continuous at π₯ equals one.
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Is π already continuous at π₯ equals one?
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Well, we have a checklist which allows us to find out whether a function is continuous at a certain point.
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π must be defined at that point.
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So in our case, we need to check that π of one is defined.
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And the limit of π of π₯ as π₯ approaches that point, in our case one, must exist.
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And finally, these two values must be equal.
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Letβs start at the top of the list.
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Is π of one defined?
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Well, weβll use the definition of π of π₯ that weβre given in the question.
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Substituting one for π₯ gives one squared plus one minus two all over one minus one.
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In the numerator, one squared plus one is two.
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And subtracting the two gives us zero.
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And in the denominator, one minus one is zero.
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So we get the indeterminate form zero over zero.
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Zero over zero and hence π of one are not defined.
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And so our function π is not already continuous at π₯ equals one.
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This isnβt necessarily a massive issue.
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After all, our task is to define π of one.
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If this is the only thing stopping π from being continuous at π₯ equals one, then we can just define π of one.
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And π will be continuous at this point, as required.
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We need to check then that there arenβt any other issues.
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We need the limit of π of π₯ as π₯ approaches one to exist.
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And does it?
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We use the definition of π of π₯ from the question.
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And of course, we know that direct substitution is going to give us an indeterminate form.
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So there must be some other way to evaluate this limit.
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Well, the factor theorem tells us that as both the numerator and the denominator are zero when π₯ is one, both numerator and denominator must have a factor of π₯ minus one.
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And indeed, we can factor the numerator to π₯ plus two times π₯ minus one.
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Doing so allows us to cancel the common factor of π₯ minus one in the numerator and denominator.
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To get the limit of π₯ plus two as π₯ approaches one.
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This is a limit that can be evaluated using direct substitution.
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Substituting one for π₯, we get one plus two, which is three.
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So yes, the limit of π of π₯ as π₯ approaches one does exist.
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And it equals three.
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Now the last thing on our checklist is that the limit of π of π₯ as π₯ approaches one must be π of one.
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We found that the left-hand side, the limit of π of π₯ as π₯ approaches one, is three.
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But the right-hand side π of one is undefined.
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But our task is to define π of one.
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If we define π of one to be three, thatβs the value of the limit of π of π₯ as π₯ approaches one.
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Then our third item on the checklist will be satisfied.
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And of course, defining π of one to be three also sorts out the first item on our checklist.
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π of one is now defined.
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So defining π of one to be three makes π continuous at π₯ equals one.
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Because π of one is now defined, the limit of π of π₯ as π₯ approaches one, as we saw, is three.
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And now that weβve defined π of one to also be three, these two values are equal.
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It might be helpful to look at the graph of π of π₯ to see what weβve done.
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As we saw with π₯ not equal to one, π of π₯ is just π₯ plus two.
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And so the graph of π of π₯ is just the straight line graph of π¦ equals π₯ plus two, with this hole here when π₯ is one.
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This hole in the graph comes because π of one is not defined.
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And as a result, π is not continuous at π₯ equals one.
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There is then this link between the technical definition of continuity and our intuitive understanding of what continuity means.
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There should be no gaps.
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We plug this gap and make the function continuous by defining π of one to be three.
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Now, thereβs no hole in the graph.
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And π is continuous at π₯ equals one.
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Letβs now see an example where we canβt simply plug the gap.
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Given π of π₯ equals one over π₯, if possible or necessary, define π of zero so that π is continuous at π₯ equals zero.
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So we have the reciprocal function here, which hopefully we know and love.
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And we know what its graph looks like.
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The graph has two pieces, one in the first quadrant and the other in the third.
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And these pieces are separated by a vertical asymptote at π₯ equals zero.
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Intuitively then, it feels like this function is not continuous at π₯ equals zero.
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Because the graph of our function is not one continuous curve but is formed of two curves with the break at that asymptote π₯ equals zero.
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Just to the left of this line when π₯ is small but negative, π of π₯ is larger magnitude but negative.
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And to the right of this line when π₯ is small but positive, π of π₯ is larger magnitude and positive.
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So as we pass π₯ equals zero, the right of π of π₯ changes from a value which is larger magnitude and negative to a value which is larger magnitude and positive.
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Letβs see if our intuition agrees with the technical definition by going through our checklist to see if the function is continuous.
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For π to be continuous at π₯ equals zero, we need π of zero to be defined.
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The limit of π of π₯ as π₯ approaches zero to exist.
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And finally, we need these values to be equal.
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Okay, so is π defined at zero?
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Well, if we substitute zero into the definition of π of π₯, π of π₯ is one over π₯, we get π of zero is one over zero.
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And this is not defined.
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Zero is not in the domain of our function.
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So π is not defined at zero.
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And hence, π is not continuous at π₯ equals zero.
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But this isnβt necessarily a surprise as our task is to define π of zero to make π continuous at π₯ equals zero.
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We wouldnβt have any work to do if π were continuous at π₯ equals zero and π of zero was defined.
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We check the second criterion.
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The limit of π of π₯ as π₯ approaches zero must exist.
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With the idea that if this limit does exist, then we can just define π of zero to be the value of this limit.
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And, then the function will be continuous at π₯ equals zero as required.
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Does this element exist?
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Well, if we look at the left-hand limit as π₯ approaches zero, π of π₯ approaches negative infinity.
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And it gets worse.
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The limit from the right is positive infinity.
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As π₯ approaches zero from the right, π of π₯ gets larger and larger without bound.
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The limit of π of π₯ as π₯ approaches zero therefore does not exist.
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And this is a problem.
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We can define π of zero to be whatever we like.
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But whatever we define it to be, this doesnβt change the fact that the limit of π of π₯ as π₯ approaches zero does not exist.
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And so π cannot be continuous at π₯ equals zero.
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So this is our answer then.
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The limit of π of π₯ as π₯ approaches zero doesnβt exist.
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And so π cannot be made continuous at π₯ equals zero by defining π of zero.
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The fact that π of zero is undefined is not really the problem here.
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And looking at the graph, we can kind of see why this is true.
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There isnβt just a small gap on the graph, which can be plugged by a single point.
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There is a huge chasm.
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While π of π₯ is not continuous at π₯ equals zero and cannot be made so just by defining π of zero, itβs worth pointing out that the function π of π₯ is actually continuous at other values of π₯.
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Given π of π₯ equals one over π₯, if possible or necessary, define π of one so that π is continuous at π₯ equals one.
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We go through our checklist with this new question.
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We need π of one to be defined.
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The limit of π of π₯ as π₯ approaches one to exist.
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And the limit of π of π₯ as π₯ approaches one to equal π of one.
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Is π of one defined?
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Yes, π of one is one over one, which is one.
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π of one is defined.
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One is in the domain of π.
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Now, does the limit of π of π₯ as π₯ approaches one exist?
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Letβs look at the graph.
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As π₯ approaches one from the left and as π₯ approaches one from the right.
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The value of π of π₯ approaches the same at finite value, the value one.
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If we wanted to find the value of this limit without using the graph, weβd probably just use direct substitution.
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We can see then that the third item on the checklist at the limit of π of π₯ as π₯ approaches one equals π of one is also satisfied.
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The values of the left-hand side and right-hand side are both one.
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Our answer then is that π is already defined and continuous at π₯ equals one.
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Itβs not necessary to define π of one again to make this so.
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Letβs see one more quick example before we conclude.
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Is the function π of π₯, which is piecewise defined to be two π₯ plus four all over π₯ plus two if π₯ is less than negative two.
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Zero if π₯ is equal to negative two.
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And π₯ squared plus six π₯ plus eight all over π₯ plus two if π₯ is greater than negative two, continuous at π₯ equals negative two.
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To find out, we go through our checklist.
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For the function π to be continuous at π₯ equals negative two, π of negative two must be defined.
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The limit of π of π₯ as π₯ approaches negative two must exist.
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And these values must be equal.
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So we start by checking that π of negative two is defined.
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Well, thatβs easy to see from the question.
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Weβre told that π of π₯ is zero if π₯ is negative two.
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So π of negative two is defined.
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π of negative two is equal to zero.
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Now, we need to check that the limit of π of π₯ as π₯ approaches negative two exists.
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There are different rules for the values of π of π₯ when π₯ is less than negative two and when π₯ is greater than negative two.
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Which weβll use to check the left-hand and right-hand limits, making sure that they exist and are equal.
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Whatβs the left-hand limit?
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Well, to the left of negative two, π of π₯ is defined by two π₯ plus four all over π₯ plus two.
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So we have to find the limit of two π₯ plus four all over π₯ plus two, as π₯ approaches negative two from the left.
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If we try direct substitution, we get the indeterminate form zero over zero.
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So we have to find this limit some other way.
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We do this by factoring the numerator, getting two times π₯ plus two.
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And this factor of π₯ plus two cancels with the factor of π₯ plus two in the denominator.
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And so our limit is that of the constant function two whose value must be two.
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So the left-hand limit exists.
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What about the right-hand one?
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To the right of π₯ equals negative two, π of π₯ is this algebraic fraction.
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And we can find the value of this limit in a similar way.
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Factoring the numerator and canceling the common factor of π₯ plus two, to get the limit of π₯ plus four as π₯ approaches negative two from the right.
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This can be solved using direct substitution.
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We get negative two plus four, which is two.
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So this right-hand limit also exists and is equal to the left-hand limit.
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Both of them have the value two.
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Hence the limit of π of π₯ as π₯ approaches negative two from either direction also exists and has the value two.
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All thatβs left to check is this third point.
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For π of π₯ to be continuous at π₯ equals negative two, we need the limit of π of π₯ as π₯ approaches negative two to be equal to the value of π at negative two, π of negative two.
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But weβve just seen that the value of this limit is two, whereas π evaluated at negative two is zero.
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The limit of π of π₯ as π₯ approaches negative two is not equal to π of negative two.
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Although the first two conditions on our checklist are satisfied, the last one isnβt.
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And so our answer is no.
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The function π is not continuous at π₯ equals negative two, as the limit of π of π₯ as π₯ approaches negative two is not equal to π of negative two.
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Letβs now conclude with the key points that weβve covered in this video.
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A function π is said to be continuous at a number π if the limit of π of π₯ as π₯ approaches π equals π of π.
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To check the continuity at π, we must check that π of π is defined.
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That is, that π is in the domain of π and that the limit of π of π₯ as π₯ approaches π exists.
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And only then does it make sense to ask if they are equal.
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So if π of π is not defined or if the limit doesnβt exist, then π is not continuous at π.
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However, if π of π is undefined, but the limit of π of π₯ as π₯ approaches π exists, we can define π of π to be the limit of π of π₯ as π₯ approaches π to make π continuous at π.
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If the limit of π of π₯ as π₯ approaches π does not exist, we canβt make π continuous at π by defining or redefining π of π.