WEBVTT
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One-Sided Limits
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In this video, weβll learn how to find the value of one-sided limits both graphically and algebraically.
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As the name implies, one-sided limits involve approaching a point, letβs say π₯ equals π, from one direction.
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That is, the positive or the negative direction.
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Now at first, it may not seem apparent why this would be useful to us.
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But letβs explore further using the following example.
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Consider the function π of π₯ is equal to π₯ plus one at all points where π₯ is not equal to two.
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If we wanted to find the limit as π₯ approaches two of π of π₯, we could do so by direct substitution.
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We can do so because even though π₯ equals two is not in the domain of our function, the limit concerns values of π₯ which are arbitrarily close to two but not where π₯ is equal to two.
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Performing a substitution, we find our answer is two plus one which is three.
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Letβs now take a look at our function πof π₯ on a graph.
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We first note that the hollow dot here tells us that the function π of π₯ is not defined at this point, where π₯ is equal to two.
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Letβs now consider the process of taking a limit but graphically.
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We know that our limit concerns values of π₯ which are close to two.
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So letβs look at a value which is just slightly smaller; letβs say 1.8.
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If π₯ is 1.8, the value of our function is π of 1.8.
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The value of this is 1.8 plus one which is of course 2.8.
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To get a more accurate limit, we need π₯ to approach our value of two.
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Letβs now look at π₯ equals 1.9.
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In this case, the value of our function is 2.9.
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We could continue this process getting closer and closer to the value of π₯ equals two.
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Doing so we find that the value of our function approaches three as we would expect.
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Here, we note that we have approached our value of π₯ equals two from the left or from the negative direction.
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We could also perform the same exercise approaching from the right or from the positive direction.
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In fact, if we were to do so we would find that our values of π of π₯ would converge to the same thing.
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The point that weβre illustrating here is that approaching the value of π₯ equals two from the left or the right seems to approach the same value of π of π₯.
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Letβs now consider what would happen if we had a different function, letβs say π of π₯, which is defined piecewise as the following.
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π of π₯ is π₯ plus one if π₯ is less than two and π₯ plus two if π₯ is greater than two.
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Again we know graphically that these hollow dots tell us that the value of π of π₯ is not defined at these two points.
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And in fact the function π of π₯ is not defined where π₯ is equal to two.
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If we we were to perform the same exercise as before approaching the value of π₯ equals two from the left, our value of π of π₯ approaches three as weβve seen.
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However, now if we approach π₯ equals two from the right or the positive direction, we would see that the values of π of π₯ seem to be approaching four.
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This means that as we move towards π₯ equals two, our values for π of π₯ seem to be approaching two different values depending on which direction weβre approaching from.
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Given this is the case, it does not make sense for us to assign a value for the limit as π₯ approaches two of π of π₯.
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And in fact we say that this limit does not exist.
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However, it is still useful for us to consider what happens as we approach from the left or the right, as this still provides useful information about our function.
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When approaching from the different directions, we have in fact found the left-sided limit and the right-sided limit for our function π.
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Now, the difference in notation here is quite subtle.
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But here we see that the negative symbol in the position that a power would usually go tells us that weβre approaching from the negative direction and the positive symbol tells us that weβre approaching from the positive direction.
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A slightly more formal definition for our one-sided limits are then as follows.
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When π of π₯ can be made arbitrarily close to some value πΏ as π₯ approaches some value π from the left, that is to say that π₯ is strictly less than π and not equal to π, then we say that the left-sided limit as π₯ approaches π of π of π₯ is equal to πΏ.
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When the same conditions are true but π₯ approaches π from the right, that is to say, π₯ is strictly greater than π and not equal to π, then we instead say that the right-sided limit as π₯ approaches π of π of π₯ is equal to πΏ.
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Here weβve seen that one-sided limits give us a useful tool for looking at functions using a piecewise function with a discontinuity as an example.
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Now letβs look at an algebraic example where one-sided limits come in handy.
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Find the left-sided limit as π₯ approaches π of π of π₯ given that π of π₯ is equal to five π₯ cos five π₯ plus two sin five π₯ over π₯ if π₯ is greater than zero and less than π over two and four over two cos nine π₯ plus π if π₯ is greater than π over two and less than π.
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Here, we have a piecewise function π of π₯ defined over two different intervals.
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Our function is clearly undefined when π₯ is less than or equal to zero or greater than or equal to π.
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We can also note that due to these strict inequality symbols π of π₯ is also undefined when π₯ is equal to π over two.
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Now, our question is asking us for the left-sided limit as we can see from this minus symbol.
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This means that weβre approaching a value of π₯ equals π from the negative direction.
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And π₯ is strictly less than π.
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Even though we know that π₯ equals π is not in the domain of our function, we can still try and find a limit since the limits concern values of π₯ which are arbitrarily close to π but not equal to π.
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Now, we know that the values of π₯ that weβre interested in are less than π but very close to this value.
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Hence, the interval of our function that weβre interested in is this one where π of π₯ is equal to four over two cos of nine π₯ plus π.
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And we can see this from our inequality.
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Since this is the interval for which π₯ is just less than π, we proceed to evaluate our limit as follows.
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We take a direct substitution of π₯ equals π into our function.
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Looking at the term cos of nine π, we recall that cosine is a periodic function and repeats itself over a period of two π radians.
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This means that cos of nine π is equal to cause of π.
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And of course this is equal to negative one.
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If we perform this substitution, we find that our answer becomes the following quotient.
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And we then arrive at an answer of four over negative two plus π.
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We have now answered the question.
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And we have found the left-sided limit as π₯ approaches π of the function π of π₯.
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In some cases, it might be difficult to sketch your function.
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And here weβve demonstrated taking a one-sided limit without plotting our function graphically.
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Itβs also worth reflecting back on the fact that since π of π₯ is undefined when π₯ is greater than π and indeed when π₯ is equal to π, then the right-sided limit as π₯ approaches π of π of π₯ does not exist.
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And indeed the normal limit as π₯ approaches π of π of π₯ cannot be said to exist either.
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In this case, it only makes sense for us to assign a value to the left-sided limit as π₯ approaches π of π of π₯ demonstrating how one-sided limits give us increased precision in our mathematical descriptions.
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In the example we just saw, values of π₯ greater than π were not in the domain of our function π.
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And hence, we said that the right-sided limit as π₯ approaches π of π of π₯ did not exist.
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However, we can now look at an example to illustrate that even if π of π₯ seems to be defined over all values of π₯, then maybe some cases whether left or the right-sided limit may not exist.
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And indeed the normal limit in this case would not exist either.
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Letβs take a look at one such example.
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Determine the left-sided limit as π₯ approaches negative nine of π of π₯ and the right-sided limit as π₯ approaches negative nine of π of π₯ given that π of π₯ is equal to π₯ plus nine if π₯ is less than or equal to negative nine and one over π₯ plus nine if π₯ is greater than negative nine.
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Here, weβve been given a function defined piecewise over two intervals.
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For the left-sided limit, weβre approaching π₯ equals negative nine from the negative direction.
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Hence, π₯ is less than negative nine.
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For the right-sided limit, weβre approaching π₯ equals negative nine from the positive direction.
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And hence, π₯ is greater than negative nine.
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Since π₯ equals negative nine is the point between the two intervals of our piecewise function, for our left-sided limit will be in the first interval and for our right-sided limit will be in the second interval.
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Letβs work on finding the left-sided limit.
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In this case, our function π of π₯ is π₯ plus nine.
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We can find this limit by direct substitution of π₯ equals negative nine into our function.
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Doing so, we find that our answer is negative nine plus nine which is equal to zero.
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The left-sided limit as π₯ approaches negative nine of π of π₯ is therefore zero.
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Now for the right-sided limit, here our function π of π₯ is one over π₯ plus nine.
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Again, we tried direct substitution of π₯ equals negative nine into our function.
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This time, doing so gives us an answer of one over zero.
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And as we know, dividing one by zero cannot be evaluated to a numerical value.
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In cases such as this, we say that the limit does not exist.
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And so in a strict sense, this is the answer to our question.
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To more fully understand our result, however, letβs look at a graphical plot of our function.
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Here, we have sketched our graph.
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And we know that, in the interval where π₯ is less than or equal to negative nine, we have a well behaved function.
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And we know that due to the solid dot here at negative nine π₯ is indeed defined at this point.
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For the other interval, we know that as π₯ approaches negative nine, we have a vertical asymptote.
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This means that the values of π of π₯ get arbitrarily large.
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And this is often represented as infinity.
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In this sense, it is common to write that the right-sided limit as π₯ approaches negative nine of π of π₯ is equal to positive infinity.
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A very important distinction to make here is that we are not saying that infinity takes a numerical value.
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Nor are we saying that our limit exists.
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Rather, weβre expressing that the limit does not exist in a particular way.
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We do this because expressing a limit like this still gives us useful information about our function, as shown by the graph.
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Writing the limit in this way even without the graph would give us a sense that, at negative nine, we have a discontinuity.
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And as we approach this value from the right, the values of π of π₯ get arbitrarily large.
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To round out the concept of one-sided limits, itβs useful to understand the distinctions and the relationships between the value of our function at π₯ equals π, the normal limit of the function as π₯ approaches π, and, of course, the left and right limits as π₯ approaches π.
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In particular, let us first formalize the following relationship.
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If the left-sided limit and the right-sided limit as π₯ approaches π of a function π of π₯ exist and are equal to each other, taking some value πΏ, then the normal limit as π₯ approaches π of the function π of π₯ also exists and is also equal to πΏ.
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In fact, we could also take this rule in reverse by saying that if the normal limit as π₯ approaches π exists and is equal to πΏ, then it will be equal to the value of the left-sided and the right-sided limits as π₯ approaches π.
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And these would also be πΏ.
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Weβve already touched on this in previous examples; but here we reiterate.
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If the left- and the right-sided limits disagree or do not exist, then it does not make sense for us to say that the normal limit itself exists.
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Itβs also worth noting that the limit of a function as π₯ approaches π may, in fact, be completely independent from the value of the function itself at the point where π₯ is equal to π.
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Letβs look at an example of this in the following question.
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Find the following: π of negative three the left-sided limit as π₯ approaches negative three of π of π₯, the right-sided limit as π₯ approaches negative three of π of π₯, and the normal limit as π₯ approaches negative three of π of π₯.
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This is part a) through d).
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Next for part e) through h), we have find π of one, find the left-sided limit as π₯ approaches one of π of π₯, the right-sided limit as π₯ approaches one of π of π₯, and the normal limit as π₯ approaches one of π of π₯.
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Now at first glance, this is a lot of work.
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But we see that the first four parts of our question are very closely related as are the second four parts of our question.
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Here, weβre gonna do things in sections, first, looking at parts a), b), c), and d).
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Now, weβve been given a graph which describes the function π of π₯.
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The first point we note is that the hollow dots on our graph describe points where π of π₯ does not exist, whereas the filled dots on our graph describe points where π of π₯ does exist.
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Looking at the value of π₯ equals negative three on our graph, we see that there is a hollow dot at the point negative three zero and a filled dot at the point negative three two.
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From this, we can say that when π₯ is equal to negative three, π of π₯ is equal to two.
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In other words, weβve just answered part a) of our question.
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And π of negative three is equal to two.
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Next, we move on to the left-sided limit as π₯ approaches negative three.
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As we approach we see from our graph that the value of π of π₯ gets closer and closer to zero.
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It does not matter here that we have a hollow dot at negative three zero since the limits concern values of π₯ which are arbitrarily close to negative three but not equal to three.
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We can then say that the left-sided limit as π₯ approaches negative three of π of π₯ is equal to zero.
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In fact, the same can be said for the right-sided limit.
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As π₯ approaches negative three from the positive direction, π of π₯ also gets closer and closer to zero.
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This means that the right-sided limit as π₯ approaches negative three of π of π₯ is also equal to zero.
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Now, given that the left- and the right-sided limits as π₯ approaches negative three both exist and are equal to the same value, we can use this general rule to say that the normal limit also exists and is equal to the same value.
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We, therefore, conclude that the normal limit as π₯ approaches negative three of π of π₯ is also equal to zero.
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We have now answered parts a) through d) of our question.
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An interesting point to note here is that even though the value of π of negative three is equal to two, the left, the right, and the normal limits as π₯ approaches negative three are all equal to zero.
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Again, the limits concern values of π₯ which are close to, but not equal to, negative three.
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In this case, the limits are completely unconcerned with the exact value of π of negative three and only concerned with the values π₯ takes as it approaches negative three.
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In fact, we could remove the point at negative three two entirely from our graph.
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And doing so would leave all of our limits completely unchanged even though the value of π of negative three would be left undefined.
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Letβs now move on to the next four parts of our question.
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First, we must find the value of π of one.
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From our graph we see that there is a filled dot at the point one, negative two and a hollow dot at the point one, four.
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We know that the filled dot is where π of π₯ is defined.
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And hence, π of one is equal to negative two.
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Next, for the left-sided limit as π₯ approaches one for π of π₯, we see what happens to our function as we approach the value where π₯ equals one from the negative direction.
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Here, itβs clear that the value for π of π₯ is approaching negative two.
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And therefore, this is also the value of our left-sided limit.
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For the right-sided limit, we see what happens to the value of π of π₯ as we approach from the positive direction.
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And from our graph, the value of π of π₯ gets closer and closer to four as π₯ approaches one from the positive direction.
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This means that our right-sided limit is equal to four.
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Now for the final part of our question, the normal limit as π₯ approaches one of π of π₯.
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Here, we know that both the left- and the right-sided limits exist.
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However, they take different values.
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And therefore, they disagree given this fact we can conclude that the normal limit does not exist.
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We have now answered all parts of our question.
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This example illustrates a few different ways in which the value of the function, the left-sided the right-sided, and the normal limits of a function can relate to each other.
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To finish off, letβs go through a few key points.
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The left- and right-sided limits of some function π of π₯ concern the value of π of π₯ as π₯ approaches some value π from the negative direction and the positive direction, respectively.
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A more formal definition has been given below here.
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If the left- and the right-sided limits both exist and agree on some value πΏ, then the normal limit will also exist and be equal to the same value πΏ.
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We can also sometimes understand this rule in reverse, making inferences about the left- and the right-sided limits given the normal limit.
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Sometimes, a limit where the normal or one-sided may be expressed as positive or negative Infinity.
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In these cases, we are not saying that infinity, whether positive or negative, takes a numerical value.
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Nor are we saying the limit exists.
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But rather, this is a particular way of expressing that the limit does not exist because it gives us useful information about the function.
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As a final point, we remember that the one-sided limits can exist when the normal limit does not.
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And they therefore provide us with useful additional tools for describing a function, for example, when looking at a piecewise function with a discontinuity.