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>> Section 1.2 Finding Limits Algebraically and Continuity. Southwestern College Professor Maria Olivas. [ Music ] Finding limits algebraically and continuity. Objectives, evaluate limits analytically or algebraically,
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determine whether a function is continuous at a point.
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One basic properties or laws of limits.
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Theorem, let's c be a real number and suppose that the limit as x approaches c of f of x equals l, and the limit as x approaches c of g of x equals m. Then, first property, the limit as x approaches c of k, where k represents a constant is going to be equal to k. So, I'm going to draw an arrow pointing at k from above, and I'm going to write down the word constant right next to the arrow. So, that way, you remember that k does represent a constant.
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And so the limit once again as x approaches c of k is just going to be equal to that same constant k. I'm going to put a little example here underneath the first property. I'm going to write the limit as x approaches c of three is going to be equal to three. The second property, the limit, as x approaches c of x equals c. Third property, the limit as x approaches c of open up a set of parentheses a times f of x,
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close the parentheses is going to be equal to a times the limit as x approaches c of f of x. Forth property, we're going to have the limit as x approaches c of we're going to open up a set of square brackets, and then we're going to have f of x plus or minus g of x, close the square brackets. And this is going to be equal to the limit as x approaches c of f of x plus or minus the limit as x approaches c of g of x. And this is equal to l plus
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or minus m. The fifth property states the limit as x approaches c of open up a set of square brackets, f of x times g of x, close the square brackets, is going to be equal to the limit as x approaches c of f of x times the limit as x approaches c of g of x.
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And this is also equal to l times m. Property number six, the limit as x approaches c of open up a set of square brackets, f of x divided by g of x, close the square brackets is going to be equal to the limit as x approaches c of f of x divided by the limit as x approaches c of g of x. And this is also going to be equal to l divided by m, provided that m is not equal to zero. Property number seven, the limit as x approaches c of open up a set of square brackets f
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of x close the square brackets raised to the nth power. This is going to be equal to we're going to open up a set of square brackets, the limit as x approaches c of f of x, close the square brackets, then raise this to the nth power. And this is also equal to l raised to the nth power provided that l raised to the nth power exists. Example 1, determine the limit the limit as x approaches one of in the numerator here we have x squared minus 5x plus 6 over x minus 3.
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So, for this particular problem, we're going to use direct substitution. So we're allowed to use direct substitution, so long as the function is not defined differently to the left and to the right of x equal to 1. And we've already worked with this function in the previous section, so we know that that's going to be true, it's not defined differently. So, we're going to use direct substitution here. I'm going to just directly substitute x equal to 1 into the function.
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First, let me rewrite the limit, the limit as x approaches 1 of in the numerator once again, we have x squared minus 5x plus 6 all over x minus 3. This is going to be equal to, as soon as I do the direct substitution I no longer have to write the limit. So, it's just going to be 1 squared minus 5 times one plus 6, all over 1 minus 3. And this is going to be equal to 1 squared is equal to 1 minus 5 times 1 is equal to 5, plus 6, all over 1 minus 3 is equal to negative 2.
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This equals negative 4, plus 6, all over negative 2, which equals 2 divided by negative 2, which equals negative 1. I'm going to go ahead and box that in. Two, algebraic strategies, functions that agree at all but one point. Strategies for determining limits. One, if possible, try direct substitution. This works with polynomials and with certain other functions as long as the function is not defined differently to the left and to the right of x equals c.
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And as long as direct substitution doesn't cause something to be undefined. Two, if direct substitution doesn't work, you can probably determine whether the limit even exists by a graphical or numerical approach. You can then estimate the limit. But better yet, three, try an algebraic or analytical approach. Use algebraic techniques such as factoring multiplying by the LCD, multiplying by the conjugate, et cetera, to rewrite the function in an alternate form.
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Example 2, determine the limit analytically. The limit as x approaches 3 of x squared minus 5x plus 6 all over x minus 3. Let's begin by using direct substitution. If I use direct substitution here, and I tried to substitute the value x equal to 3 into this function, let's see what happens. I'm going to get 3 squared minus 5 times 3 plus 6, all over 3 minus 3. This is going to give me 9 minus 15, plus 6 all over 0. And if I were to continue to work out the numerator part,
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I would end up with zero over zero. So, this is undefined. So, we cannot conclude anything from this. We cannot figure out what the limit is. I need to use another method. So, I'm going to rewrite the limit here, the limit as x approaches 3 of x squared minus 5x plus 6, all over x minus 3, and I'm going to use factoring. I'm going to factor out that numerator. So, I'm going to write the limit as x approaches 3 of the factors of that trinomial
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and the numerator are x minus 2 times x minus 3, all over x minus 3. The factor of x minus 3 will cancel out because those are a common factor between the numerator and the denominator. So, they will divide out. I will write once there in place, since they're divided out. And we're going to be factors of ones in their place. I'm going to have the limit as x approaches 3 of inside parentheses, I'm going to write x minus 2.
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Now, I can use direct substitution so I will write 3 minus 2 equals 1.
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So, I was able to find this limit analytically, or another way to say this, I found the limit algebraically. If f of x equals x squared minus 5x plus 6, all over x minus 3, and I just put a cloud around that entire function there and arrow pointing down here saying that x cannot equal 3 in this particular function. Then f of x equals g of x. And in this case, g of x is going to be equal to the x minus 2 factor for x equal to 3.
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Example 3, determine the limit analytically, the limit as x approaches zero of in the numerator here, we're going to have the square root of x plus 2 all underneath the square root minus the square root of 2, all over x. Once again, let's try direct substitution and let's see what we get. If I use direct substitution here, I'm going to substitute x equal to zero into the function. So, we will end up with the square root of zero plus 2, minus the square root of 2 all over 0.
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And if I were to continue this, I would end up with the square root of 2 minus the square root of 2 over 0. And this would ultimately give me zero over zero. This is undefined, right? So, actually, before I write that, zero over zero, let me just lasso that all my work that I did there. Let me shrink it so I have more space. Now, I'm going to write this equal to zero over zero and of course, this is undefined. So, we need to find another method to find this limit.
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So, let's try finding the limit analytically now. The limit as x approaches zero of -- and I'm going to just recopy everything, the square root of x plus 2, minus the square root of 2 all over x.
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For this particular problem, we're going to try multiplying by the conjugate, the conjugate of the numerator. So, I'm going to multiply by the conjugate of the numerator to both the numerator and the denominator of this rational expression. So, I'm going to put parentheses around my numerator, and then I'm going to multiply by -- I'm going to open up a set of parentheses, the square root of x plus 2 plus the square root of 2, close the parentheses.
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And the same thing to the denominator, I'm going to multiply by open up a set of parentheses square root of x plus 2 plus the square root of 2, close the parentheses. I'm going to use the box method to multiply what we have in the numerators. Across the top of the box, I'm going to write the square root of x plus 2 minus the square root of 2. And alongside the left-hand side of this box, I'm going to write the square root of x plus 2 plus the square root of 2.
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So the square root of x plus 2 times the square root of x plus 2 is x plus 2. Here, we're going to get negative square root of 2 times the square root of x plus 2. Over here, we're going to get the square root of 2 times the square root of x plus 2. And then, the square root of 2 times a negative square root of 2 is just staying negative 2. I'm going to look for like terms that I can combine. So, I notice here that I have a positive 2, which I'm going to circle
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and a negative 2, those will cancel out. So, I'm going to go ahead and draw a line that goes through them to show that they cancelled out. I'm also going to circle what I have alongside this other diagonal here. These expressions are going to cancel out, so I'm going to go ahead and draw a line through those expressions. And the only thing that's left is this x, which I'm highlighting in yellow. So, let's go ahead and rewrite what we have here.
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Before I rewrite what I have here leftover from my limit, I'm going to lasso all of the work that I did for the direct substitution part. I'm going to shrink it, and I'm going to raise it slightly up higher so that I have more space. OK. So, this limit is going to be equal to the limit as x approaches zero of, we're going to have x in the numerator over x times, I'm going to open up a set of parentheses, the square root of x plus 2, plus the square root of 2 and close the parentheses.
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So, I notice here that my x's are going to cancel out because they have a common factor in the numerator and in the denominator. So I'm going to highlight those x's in yellow, and I'm going to cross them out to say divide out and turn into factors of one. So, this is going to be equal to the limit as x approaches zero of 1 over the square root of x plus 2, plus the square root of 2. So, now, I'm going to go ahead and do direct substitution. And this is going to be equal to 1 over the square root
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of zero plus 2, plus the square root of 2.
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This is equal to 1 over the square root of 2 plus the square root of 2. And that's going to be equal to 1 over 2 square root of 2. And I'm going to box that in, that's going to be my final answer for that one.
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Example 4, determine the limit analytically. The limit as x approaches 1 of 1 over x minus 1 over 4, all over x minus 4. So we're going to go ahead and solve this. If we were to try to do direct substitution here, it would end up giving us something that is undefined. So, we're just going to go straight to the finding the limit here by doing analytical methods, the limit as x approaches 4 of, I'm going to rewrite this problem, 1 over x minus 1 over 4, all over x minus 4.
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So, here, I'm going to go ahead and get a common denominator here in the numerator. I'm going to go ahead and multiply the first fraction by 4 to the numerator and denominator and the second fraction in the numerator by x to the numerator and denominator. This is going to equal the limit as x approaches 4 of, and the numerator, we're going to have 4 over 4x minus x over 4x, all over x minus 4. And now, this will equal the limit as x approaches 4 of in the numerator,
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we're going to have 4 minus x all over 4x and that's going to be over x minus 4. I'm going to highlight the x minus 4 in the denominator, and I'm going to write it over a 1. Now, we're ready to multiply by the reciprocal. So the limit as x approaches 4 of, I'm going to open up a big set of parentheses here, I'm going to have 4 minus x over 4x times 1 over x minus 4, close the parentheses. Now, if we look at these two factors here, they almost look the same,
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the 4 minus x is almost the same as x minus 4. All we need to do is just factor out a negative, and that's what I'm going to do. I'm going to factor out the negative from that 4 minus x, and I'm going to get negative times open up a set of parentheses negative 4 plus x, close the parentheses. So, then we get this equal to the limit as x approaches 4, we're going to open up a set of parentheses. We're going to have a negative open up another set of parentheses,
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x minus 4, close the parentheses over 4x. This is going to be multiplied with another fraction here, 1 over x minus 4, and then the x minus 4s will cancel out with the factors of one they're in place. And then, we will have the limit as x approaches 4 of negative 1 over 4x. We can use direct substitution out. So, this is going to equal negative 1 over 4 times 4, which will equal negative 1 over 16. And I'm going to go ahead and box that in and that's going to be our final answer.
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So, we're ready for example number five, consider the following function. Consider f of x equals negative 3x plus 1 when x is less than negative 1, six, when x is greater than or equal to negative 1, and x is less than or equal to 2, x plus 4 over x minus 1when x is greater than 2. I'm going to go ahead and sketch a graph of this piecewise defined function. So, here is the y axis, here is the x axis. I'm drawing this coordinate plane here in the upper right-hand corner of the screen.
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I'm going to go ahead and draw some tick marks here on the axis. I'm going to go ahead and draw some tick marks here on the y axis. So the tick marks, I'm going to draw on the y axis are going to go from negative 2 to 9 by a scale of 1. And the ones I'm going to draw on the x axis are going to go from negative 6 to 7 by a scale of 1. I'm going to go ahead and highlight the first piece of this piecewise function. I'm going to highlight with this f of x equals negative 3x plus 1
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when x is less than negative 1. So now, I'm going to write in the upper left-hand corner of the screen, f of x equals negative 3x plus 1. So, in order to draw this sketch of this graph, I'm going to go ahead and find what f of negative 1 is equal to. F of negative 1 is equal to negative 3 times negative 1 plus 1, f of negative 1 is equal to 3 plus 1, so f of negative 1 is equal to 4. This is going to represent a hole at negative 1 comma 4, because there is no equal sign in this inequality.
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It says x is strictly less than negative 1, but I know at least that I'm going to be drawing an open circle at negative 1 comma 4. And then all the values that are less than negative 1, it's going to be a line that looks just like f of x equals negative 3x plus 1. So, I've g1 ahead I've drawn my open circle at negative 1, comma 4. I'm writing down the slope of this line, the slope is negative 3, which I'm writing as positive 3 over negative 1, so that means I'm going to go
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up 3 units, and then to the left, 1 unit. So, from my hole, I went up 3 units and to the left 1 unit, I drew a point. I connected my point and my open circle, and I extended that line to the ends of a coordinate plane. I drew an arrow at the end. So that's going to be that first piece of my piecewise defined function. Well, I'm going to highlight in green, the f of x equals 6, when x is greater than or equal to negative 1, and x is less than or equal to 2. So let's go ahead and graph this part.
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This is going to look just like y equal to 6, a horizontal line at y equals 6. So, I've g1 ahead and numbered my little tick marks up from 1 to 6 on the y axis. I've gone ahead and drawn a point right above the open circle I drew earlier at negative 1 comma 6, and I'm also going to draw a point at 2 comma 6. I'm going to connect those two points with a line segment. And then I'm going to label each one of those point with its corresponding ordered pair.
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So, I labeled first the point that was on the right of that line segment, which is 2 comma 6, the one on the left was negative 1 comma 6, and then my hole was at negative 1 comma 4. And then, the point that I use to help me draw my first piece here that was connected to my hole or my open circle, that's going to be the point that at negative 2 comma 7.
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Now, we need the third piece. The third piece says f of x equals x plus 4 over x minus 1. I'm going to highlight that piece in blue as well as the inequality that represents the domain over which this is defined. So, I've gone ahead and highlighted the x plus 4 over x minus 1 in blue and the -- when x is greater than 2 in blue, as well. First, I'm going to erase all the work I did earlier, all of the work that is
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in the upper left-hand corner of the screen, so that I have space to do my new work. I'm going to go ahead and write down here, f of 2 equals 2 plus 4 over x -- 2 minus 1, I should say, which equals 6 over 1, which equals 6. So, I know that the inequality is strictly greater than 2 for that third piece, but I'm trying to find the open circle where the ordered pair is going to be where it's located at. So, I can tell here, there's going to be a hole at 2 comma 6.
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So, I'm going to go ahead and draw my open circle at 2 comma 6. Well, actually, I don't have to draw an open circle because it happens to overlap right over a closed circle from the earlier piece from my piecewise defined function. So, I'm just going to go ahead and draw the curve that connects to that piece, and just bring it on downward and towards the right. And you can get this form by looking at the graph on your graphing calculator to get the sketch of this function.
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So, I'm going to go ahead and erase the work that I did in the upper left-hand corner of the screen so that way we can focus on finding the limits. We're going to start off with limit the part A here, the limit as x approaches negative 1 from the left of f of x. When you're approaching negative 1 from the left, remember, if you're looking at the values where x is going to be less than negative 1.
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So, let's look at our graph. I'm going to go ahead and highlight on the graph on the x axis as I'm getting closer and closer to negative 1 from the left, what value am I approaching on the function. I'm going to highlight in yellow as well on the function as I'm getting closer and closer to the x value of negative 1. And I've g1 ahead and seen that I'm getting closer and closer to 4. So I've circled the 4 on the y axis and I've set this limit equal to 4 and I've boxed
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in that answer, B. The limit as x approaches negative 1 from the right of f of x. So now, we are going to look where are we getting -- as we're coming in from the right-hand side. So those are the values that are greater than negative 1. So, I'm going to go here and write x is greater than negative 1, just to remind you, as we're coming in from the right. Those are the values that are greater than negative 1. So now looking here at the graph, I've highlighted first words as the limit words,
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as x approaches negative 1 from the right, I've highlighted that part in green. And then, in green, I've also highlighted as I'm coming in from the right-hand side closer and closer to negative 1 on the x axis and on the function. And I am approaching the y value of 6. So, I've set that limit equal to 6 and I've boxed in that answer. C, the limit as x approaches negative 1 of f of x. Well, if we look at Part A and Part B, our right-hand limit was not the same as the left-hand limit.
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These 2 values were not equal to each other. And in order for a limit to exist, the right-hand limit must equal the left-hand limit. So I've g1 ahead and I've circled where it says x is approaching negative 1 and I have drawn arrows pointing out the 2 values from the left-hand limit and the right-hand limit, and I wrote these values are not equal. So, therefore, the limit as x approaches negative 1 of f of x equals does not exist, or DNE and I've box that in.
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D, limit as x approaches 2 from the left-hand side of f of x. I'm going to erase the green highlighting I did on the function, and I'm going to highlight in blue. I'm going to highlight in blue coming in from the left-hand side on the function towards x equal to 2 and I'm going to highlight the 6 that's the y coordinate for that ordered pair 2 comma 6. So, it looks like the limit as x approaches 2 from the left of f of x is equal to 6.
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Now, I'm going to highlight coming in from the right-hand side towards x equal to 2 on the function. And as we come in, we see we're also approaching 6 for the y value. So I'm going to write at the limit as x approaches 2 from the right-hand side of f of x is equal to 6. Now, for f, if you notice it says here the limit as x approaches 2 of f of x, well, if we look at the left-hand limit and the right-hand limit, these values are both equal to each other.
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If these values are both equal to each other, then that means that this limit has to also equal 6. So, I've g1 ahead and I've circled words as x approaches 2 on the limit for f. My brown arrow is pointing to those values for the limits for d and e. And so, now I've written the limit as x approaches 2 of f of x is equal to 6 and I boxed that in.
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Section 3 here, continuity. We've noticed that some limits can be found with direct substitution while others cannot. This has to do with a concept called continuity. Informally, a function is continuous on an interval, if it can be drawn without lifting your pencil that is, if it has no gaps, holes, jumps, et cetera. Formally, a function f is continuous at c if the following 3 conditions are met. One, f of c exists. Two, the limit as x approaches c of f of x exists.
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And if this limit exists, then that means that the limit as x approaches c from the right is going to be equal to the limit as x approaches c from the left. So, I'm going to write down here, the limit as x approaches c from the right of f of x is equal to the limit as x approaches c from the left of f of x. Three --
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-- the limit as x approaches c of f of x is going to be equal to f of c. So now, that we have covered these three conditions, we are going to move on and we're going to see how these three conditions are required in order for a function to be continuous. We're going to be drawing some pictures in order to help us visualize these three conditions. All right. So, let's begin with a picture that fails step 1. So, let's recall what is step 1.
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Our step 1 or our first condition that we just talked about in order for a function to be continuous is that f of c exists. So, there must be a closed circle at the ordered pairs c comma f of c, that point must exist. When this case, it's going to fail that step. I'm going to draw a coordinate plane here on the left-hand side of my screen, and a line with a positive slope and a positive y coordinate for its y intercept, going through, and we're going to focus primarily in quadrant I.
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I'm going to erase a piece of this line right here in quadrant I and I'm going to draw an open circle right there. I'm going to go ahead here, and I'm going to draw an arrow on the right-hand side of this hole pointing at that hole. And I'm going to label this as a hole, this open circle, that's what it represents. I'm going to draw a dashed vertical line going down from the hole to the x axis, and I'm going to label that with the x value of c. So this hole, this label here,
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and its location is that the ordered pair of c comma f
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of c. Another possibility would be the second coordinate plane that I'm drawing on the right of this graph. I've drawn another coordinate plane. I've drawn a dashed line that represents a vertical asymptote at x equal to c, and I've labeled it with its equation below. Now, I am drawing a function that is going to be asymptotic with this vertical asymptote that I just drew. So, I'm drawing a curve on the left-hand side and a curve here on the right-hand side.
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And the closer and closer they get to x equal to c, they are just going to become asymptotic with that vertical asymptote, and these are going to also be asymptotic as they tend to negative infinity and positive infinity would be x axis. So, these are examples of how f of x is not continuous at x equal to c. And the reason it's not continuous at x equal to c is because f of c does not exist. So now, we're going to go ahead and move on to our second picture that we're going to draw here.
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It's going to pass step 1 in this case, but it's going to fail step 2. If you recall step 2 or condition 2 stated that the limit as x approaches c of f of x exists, that means that the right-hand limit is equal to the left-hand limit. So, for our condition 1, for in order for a function to be continuous f of c exists, we're going to make sure that it passes that step. Step 2, the limit as x approaches c of f of x exist, this is the second step. We're going to make sure it fails this step.
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So, I'm going to put a red x next to this one. I put a green checkmark next to here, f of c exists but now I'm going to put a red x to make sure it fails this step. I'm going to draw a coordinate plane here on the left-hand side of our screen. And I'm going to make sure that I draw a function where the right-hand limit and the left-hand limit are not equal to each other.
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Let me begin by labeling on the y axis the value of m. So, I'm going to draw a tick mark here where it's -- where represents m for y. I want to draw a dashed horizontal line segment going out to the quadrant I. And I'm going to draw a solid close circle point. And I'm going to draw a line or a ray coming from out from that point with a positive slope. And so, it's going upwards and towards the right. And from that same point that I drew, I'm going to draw a dashed vertical line segment going
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down to the value of c on the x axis. So, they're in between that solid point and my tick mark that I drew here for c on the x axis, I'm going to draw a hole in open circle. And from that open circle, I'm going to draw a line that extends out in the opposite direction or a ray that extends out in the opposite direction. So, this line, or ray is going to go downwards, and it's going to go to the left.
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So, for step 1, it passed it, f of c is equal to m. Step 2, the limit as x approaches c from the left-hand side of f of x. This is not equal to the limit as x approaches c from the right-hand side of f of x. So, since these two limits are not equal to each other, this means that the limit as x approaches c of f of x does not exist. So, I've drawn an arrow from this, what I wrote here for the second step here where I put at the right-hand limit did not equal the left-hand limit,
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pointing up towards our step 2, then I erased it. And I drew another little arrow right underneath where I wrote the right-hand limit does not equal the left-hand limit and I'm writing the limit as x approaches c of f of x does not exist. I'm going to go ahead and I'm just going to lasso my work here and I'm going to just shrink it slightly, just so that I have a little bit more space here so it looks like it's not too overly large.
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And I'm also going to take the steps that I wrote here at the top where it wrote f of c exists, and the limit as x approaches c of f of x exists, but that 1 did not pass, right? I put a red x next to it and I shrunk those a little bit. And to the right-hand side here, I'm going to go ahead and write out.
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This is an example of how f of c is not continuous at x equal to c. And the reason it's not continuous this time is because it fails the second step, the limit as x approaches c of f of x does not exist.
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OK. Let's go on to our third example here that we're going to draw. We're going to draw an example where it passes step 1 and step 2, but it's going to fail step 3. If you recall, step 3 is when f of c is not equal to the limit as x approaches c of f of x. So for the first step f of c exists, green checkmark, yes, it's going to pass that step. We're going to make sure that our picture passes that step. Step 2, this is the limit as x approaches c of f of x exists.
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So, yes, we're going to make sure it passes this step, that means that the right-hand limit is going to equal the left-hand limit. So I'm going to write down here for 2, the limit as x approaches c from the left of f of x is going to equal the limit as x approaches c from the right of f of x. And I'm going to put a green checkmark next to that one because, yes, we're going to make sure it passes that step. This means the limit as x approaches c of f of x equals l.
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So, green checkmark next to that one. Number 3, this is the 1 step it's not going to pass, f of c equals the limit as x approaches c of f of x. We're going to make sure it doesn't pass this step. So, I'm going to put a red x here for this one, right on top of the hole thing that I just wrote out here, because it's not going to pass that step. I'm going to go ahead and draw a coordinate plane here on the left-hand side of the screen. I've labeled the value of x here on the x axis, I called it c
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and I've drawn a little tick mark there. And then, I've also labeled a value for y on the y axis, and I call that m, draw a little tick mark, and I drew a point that is the -- it's going to have the ordered pair c comma m. Then I have drawn a line that goes through here, the line has a positive slope, and it's got a positive y-coordinate for our y-intercept. I have drawn a hole or an open circle on this line right below my point that I drew
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at c comma m, and the y value for that hole or that open circle is going to be at l. So here, we're going to have 1, f of c is equal to m. So yes, f of c does exist. Two, the limit as x approaches c of f of x is going to be equal to l. So, the limit exists. And three, this condition is not met, because m is not equal to l. So therefore, it does not represent a continuous function. F of c is not equal to the limit as x approaches c of f of x.
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So I'm going to put a green checkmark next to the one, green checkmark next to the two, and then number three, well, I kind of that kind of states what we tried to accomplish here where we tried to make sure it did not pass that third step. So, I'm just going to draw an arrow from step 3 up to the first step 3 that I drew up at the top where I drew my red x. And I'm going to write right here underneath. This is why this failed, step 3. In other words, our function here is not continuous.
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So OK, we're ready to move on to some examples, where we actually demonstrate whether or not a function passes these conditions. In order to prove something is or is not continuous at a point you must use this three step formal definition. You may state without proof that polynomial functions are everywhere continuous. Note that points of discontinuity may be related to restricted values of the domain, but are not necessarily the same.
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Example 6, determine whether the function is continuous at the given point. So, for A we have f of x equals x squared minus 4 over x plus 2, when x equals 5. Step 1, f of 5 is going to be equal to f squared minus 4 over 5 plus 2. So, this is going to give me 25 minus 4 over 7, which ends up giving me 21 over 7, which simplifies to 3. So that tells me, yes, it passes step 1, f of 5 exists. Let's check step 2 and let's see if the limit exists when x approaches 5.
40:11
So, for step 2, I'm going to write the limit as x approaches 5 of x squared minus 4 over x plus 2. So, a quick way to check to see if this limit exists is to make sure you just go ahead and put this function into your graphing calculator and make sure that the right-hand limit and the left-hand limit is equal to each other at x equal to 5. And then you could just substitute 5 here as long as 5 is not a restricted value from the domain.
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And if we do that, we are going to get that this limit is equal to 3. So, yes, it works. Step 3, f of 5 has to be equal to the limit as x approaches 5 of this function. So, let's see f of 5, yes, it is equal to the limit as x approaches 5 of x squared minus 4 over x plus 2, because we got the same value of 3 for both of these. So, yes, it passes Step 3, I've put checkmarks next to each one of those steps to show it passed, because 3 is equal to 3.
41:11
And so therefore, f of x is continuous at x equal to 5. OK. Let's go on to B. For B here, it says f of x. First, let me draw a vertical line segment here separating the work from A and B. B states f of x equals x squared minus 4 over x plus 2 when x is equal to negative 2. So, first, let's check and see if f of negative 2 exists. So step 1, f of negative 2 is going to be equal to negative 2, we're going to put that inside parentheses, and we're going to square it minus 4 over negative 2 plus 2.
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So, this means f of negative 2 is going to be equal to 4 minus 4 over negative 2 plus 2, which is equal to zero over zero. Unfortunately, here f of negative 2 does not exist because this is undefined. So, we're just going to write here f of negative 2 does not exist. So, therefore, it does not pass step 1, and f of x is not continuous at x equal to negative 2.
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So, these were our examples to check to see if a function is continuous at a given point. [ Music ]
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