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0:05
BAM!!! Mr. Tarrou. In this video we are going to define continuity. Now you might have been introduced to continuity of functions in PreCalculus. We are going to expand on that idea and give you of definitions and properties that you should be getting introduced to in your Calculus class. We are just going to expand that knowledge if you got to this in your Precalculus book. If not, we are going to define it and show you some examples that hopefully you will
0:31
get complete understanding whether you have seen this concept or not before. The continuity of a function... Now we are going to in this video define continuity on open and closed intervals, and show how one sided limits allows us to check for continuity on closed intervals. I will be showing you the properties of continuity. I will be defining the two different types of discontinuity which are removable and non-removable. Then we will finish up by doing five examples,
0:55
all with different settings, that hopefully will help you really completely understand this concept and of course complete your homework and understand it. A function 'f' is continuous at 'c' if three conditions are met. We are talking about right now about continuity at a point. Now the value of c here that we are talking about is an independent variable. Some value that you can plug into the function. And again, continuity at a point. We need
1:21
to make sure that the function at c, that f(c) is defined. That I can even plug it in and get an answer. I am not trying to plug c in and it is not a value where there is a hole in the graph, or a vertical asymptote as an example. Those are very common places of discontinuity. The limit as x approaches c of f(x) exists. So we need to have a real limit now. I don't know if in your textbook if you have been formally introduced to infinite
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limits or not. But we are talking about real limits. We are not talking about a graph going off to infinity at a particular value of c. And the limit as x approaches c of f(x) is equal to f(c). Now this is very complicated looking and there is three steps that we need to go through to check for continuity of a function. But visually continuity is very simple. It is just simply can you take you pencil, put on your paper, and draw the entire
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function in a continuous form. (does not need to be smooth) A discontinuity is anywhere you have to pick up your pencil and maybe skip or draw a hole in the function, or where you have to skip a vertical asymptotes as examples So if you can draw a function and never have to pick up your pencil, then on that interval would call it continuous. But this is what it looks like mathematically. Continuity on an open interval. A function
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is continuous on an open interval (a,b) if it is continuous at each point in the interval. Now remember we are using proper interval notation. This is not coordinate a,b. So your interval notation you started hopefully to use in Precalculus. Talking about square brackets and parenthesis. The parenthesis are your open interval. If you are not sure what I am talking about... at least after I get this one last definition. Everywhere continuous
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is continuous on the entire number line from negative infinity to positive infinity. Think of your sine function as just a wave from negative infinity to positive infinity. Or a parabola that is smooth and continuous and extends in both ways forever. Or a line which is not vertical. Examples of discontinuity again at a point. Discontinuities at 'c'. Well the function could be undefined. Now there is three ways for... well I have drawn
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three ways. I have actually three ways... I have actually kind of only drawn two really... where the function is undefined. Maybe there is a hole in the graph. Holes, at least in the PreCalculus level and maybe in more difficult settings later on, but our experience of finding holes is with rational functions. Where if you have a fraction, or a rational function, and there is a factor in the numerator and the denominator that can cancel out and simplify
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that function... well that factor that can be cancelled out, if one exist, you would still need to account for that x that made the original function undefined. There you get a hole. Vertical asymptotes can also come from rational functions where the denominator is undefined but it does not cancel out. So these are examples of the function being undefined. Thus I would have to pick my pencil up to draw these graphs. I could not do it in one
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motion, so it is discontinuous at c. The limit as x approaches c of f(x) does not exist. Don't forget we are talking about real limits, it is true for infinite limits as well, you must be approaching the same value from both sides. So this graph does not exist at c, but there is a limit at c. It is whatever that y value is. Maybe it is approaching 4 from both the left and the right. It is getting to that 3.99999 if you remember finding your
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limits numerically and graphically. Here you are approaching a different y value from the left and the right. So the left and the right hand limits while they exist for this function are not equal. So the overall two sided limit does not exist. Then finally... I said in my example here... I wrote that these were equal. But actually I need to show that they are not equal. You know a function can have a limit and be defined somewhere else. So
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here our function has got a left and right limits that are equal. Maybe it is approaching a y value of 3, but there is another portion of the function that says at f(c) this function is equal to some other value... just a random specific point that is defined. If the function value and the limit value are not equal, again I am going to need to pick up my pencil as I draw this picture, it is not continuous at c. Going back over here this has a limit
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at c, though the function does not exist. This function, we will be formally introducing infinite limits very soon. The function is approaching positive infinity from both the left and right of c. So this function would very most likely have in infinite limit, or the limit as x approaches c of f(x) is positive infinity. Whereas this vertical asymptote the graph approaches positive infinity as it approaches c from the left and negative
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infinity as it approaches c from the right. Thus this has no limit, not even an infinite limit. Ok. At least not a two sided limit. Well. I am going to define this on the next screen anyway, but these holes are going to be defined as removable discontinuities. The asymptotes where there is a sharp break and you just cannot fix that, those are going to be called non-removable discontinuities. Let's get those definitions. This page is
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just some definitions that if you need in your notes maybe you want to pause the video and write them down. But I will just run through these relatively quickly. The two types of discontinuity are removable and non-removable. Our removable discontinuities, f can be made continuous by appropriately defining, or redefining f(c). Basically it is just saying there is a hole here. And if I just make a special addition to my function, make it a piecewise
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function if you will, and say maybe my limit as x approaches c of f(x) is 5. But the function itself does not exist there. So let's just add a special case, an extra line on my function that says f(c) is equal to 5. I am going to defined the function to equal the limit as x approaches c of the function filling the hole and make the function continuous. A non-removable discontinuity is that clear break, or it is not just a simple hole. The real limit to
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the left and the right of c are not equal. That brings me over to the right hand side of my board here, that the limit of the function as x approaches c from the left and right are not equal. You can't just fill in a hole and go, oh I can now draw this without picking up my pencil. The One-Sided limits and Continuity on a Closed Interval. Now we are going to show how to check for continuity on a closed interval on the next screen, but let's first
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make sure that we understand left and right limits and how they define whether a limit exists. And I have been stressing the difference between a real and an infinite limit because coming out of a PreCalculus you might believe that the only type of limits that exist are real limits. That the function is approaching a real value from the left and the right. There is such a thing as an infinite limit where it, in layman terms, it approaches either
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positive infinity or approaches negative infinity from both sides of your value of c. So the limit from the right, the limit as x approaches c from the right of f(x) is some capital L... a real y value. The limit from... it will usually be a y value... The limit from the left is x approaches c from the left of f(x), if we are plugging in x's we are getting y values, is equal to capital L. When the left and right hand limit are the same capital
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L, the same real value, then if they are equal the existence of a really limit exists. Let f be a function and let c and L be real numbers. If the left and right limits are equal then the limit as x approaches c of f(x) is equal to L. What is the difference here? From the left, from the right, now no little exponent... no sign on the upper right hand corner of the c. That means that we are talking about a two sides limit. Let's see how those one
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sided limits allow us to check for continuity on a closed interval. Little bit more explanation before I get to those five examples. Definition of Continuity on a Closed Interval. A function f is continuous on the closed interval [a,b] if it is continuous on the open interval of (a,b) and you did not forget that interval notation you started using in PreCalculus hopefully. Here we are not including a and b in the interval and here we are. It is like
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the difference between greater than and greater than and equal to. The limit as x approaches a from the right of f(x) equals f(a). That means that you are continuous from the right going towards the left side of your closed interval. And the limit as x approaches b from the left of f(x) equals f(b). So basically what does that mean? It looks and sounds really complicated. But we have a function which is continuous on the open interval and the
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only difference here is between continuous on an open and closed interval is I have to make sure that the functions value is the same as the limit values. Like say if this is an open dot and this is a closed dot down here, the function is taking on a different value at b... A different value from the limit as x approaches b from the left. Well that is not going to be continuous on the closed interval. But again, if at the endpoint of
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each... the left and right side of the interval, if the function's value is the same as the value of the limit, then we can say that it is continuous on the closed interval as well as the open interval. Properties of continuity. If b is a real number and f and g are continuous at c, then the following functions are also continuous at c. Scalar multiples like do you have a function that is continuous at c, then simply multiplying it by two will
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give you another function... a scalar multiple that as well be continuous at c. Sum and difference, if f is continuous at c... Say if f(x)=x and f(x)=sin(x) is continuous at c, then if I add or subtract those two functions that final difference function in that case will also be continuous at c. Product, the same thing. If this linear function is continuous at c, and this linear function is continuous at c, then when I multiply them together I am
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going to get another product function that is continuous at c. The same thing holds true for quotients. Just make sure that as you check for continuity for your quotient, you cannot be dividing by zero. So the function g, or your denominator, cannot be zero of course at this value of c. That would not even be part of the domain. The following functions are continuous at all points in their domain: Polynomials, think like lines
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and parabola and cubic functions. Rational functions, you cannot divide by zero... but everywhere else or for all values that are in the domain of that rational function, that rational function will be continuous along that entire domain. Radicals and trig functions as well. If you are in these function's domain, then those functions are going to be continuous along all of those values. You can check is very simple, it is continuous. Now what does
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that really mean. That means that when I give you a function and say either check for continuity at a particular point or find values at which the function is discontinuous, much of your work is going to be focused on finding values where that function is undefined. Or in piecewise functions testing where the breaks are between those parts or pieces of your piecewise function. That is what you are going to see now that we are finally getting to our five examples.
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So I have six examples instead of five. I miscounted it. Now you don't have any problems finding limits then you can skip these two examples. But you cannot test for continuity if you are not exactly sure how to find a limit. So I am going to find the limit as x approaches four from the right. And I am going find this for the rational function. But this rational function is undefined at four. If I take four and plug it in, I am
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going to get 4-4 is 0. So if you did not pick it up on your previous section of finding limits analytically, or basically with algebra techniques, then let's remind you. I need to try and manipulate this function some how so that I can come up with another function that will agree with this function everywhere except for at this value of c or 4. Because this function is undefined at 4. What I am going to do, is I am going to multiply the
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numerator and the denominator by the square root of x plus 2. We are going to get the limit as x approaches 4 from the right... the square root of x times the square root of x is x. The square root of x times positive 2 is positive 2 square root of x. Now we have -2 square root of x. Finally -2 times positive 2 is -4. That is going to be over... Now I am not going to multiply the factors in the denominator together because I want that factor
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to cancel out so I am not dividing by zero. (x-4) times the square root of x plus 2. Well what is going to happen? Thankfully some nice stuff is happening. We can do 2 square root of x minus 2 square root of x is equal to zero. So now my numerator is going to be x-4. I hope you are trying to scream through the camera, STOP you have a cancellation. That is exactly what is going to happen. I am just showing all of my work for your understanding.
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And if I do make a careless mistake somewhere and do you know... 4+4=9... then I can find that mistake in my work and either learn from it or easily fix it and redo the rest of my problem. That common factor in the numerator and denominator are going to cancel out. x -4 divided by x-4 is equal to 1. I still have that 1 in the numerator. Now I have another function that agrees with the original function everywhere except for the value of 4. We are
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going to plug it in and find the limit. That is equal to 1 over the square root of 4+2. That is going to be equal to 1 over 2+2, which is equal to 1/4. So the limit as x approaches four from the right of this original function is equal to 1/4. If this rationalization of the numerator had not created another function in which I could directly substitute in the value of 4, then I would be forced to go back here and find the limit numerically or graphically
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and see what the function's limit is approaching as x is approaching 4 from the right. Over here we have the limit as x approaches 3. There is no sign in the exponent area on the three so we are talking about a two sided limit here. I want to know if there is a real limit for this function at 3. We have 3 plus... this is a greater integer function of x. The greater integer function says that I want... the value I get from this term is greatest
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integer n which is less than or equal to x. So if I want to draw a graph of this and I say let x equal .2... well .2, I am plugging in .2 but I have to get out an integer and it must be less than or equal to x... So if I plug in .2 I am going to get out a value of 0. So for x values of 0 up and till I get to 1, this is going to be equal to zero and 3+0 is equal to 3. So we have a closed dot, we are going to have an open dot where we
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have x=.9999999 If I am plugging in .999999 and I am supposed to get out an integer that is less than or equal to x, that is going to come out to be 0. But if I plug in 1, I am going to get out an n value of 1 because it is equal to x. So when I get to 1, I am going to have a closed dot up here and go until an open dot at x=2. Again step up one to a... When I plug in 2, put in a 2... get out a 2 and 2+3 is equal to 5. Plug in 2.1,
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get out 2, 3+2 is still equal to 5. We go to an open dot at the 3. As soon as I plug in the exact value of 3 I am going to step up to six. So we can see by the graph that as I find the limit as x approaches 3 from the left of 3 plus the greatest integer of x, that is going to be equal to... approaching 3 from the left... and that is going to give me a left limit equal to 5. The limit as x approaches 3 from the right of 3 plus the
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greatest integer of x... approaching 3 from the right... approaching 3 from the right... that is supposed to be between 3 and 4... as I approach 3 from the right my limit is 6. My left and right hand limits are not equal, thus the limit as x approaches 3 of f(x) does not exist. Now again, we need to start using this idea like here if the limit does not exist then I know that this function is not continuous at the x value of 3. That is why
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I am reviewing with you how to find limits before we get to our next four examples where we do check for continuity or look for places where the function is discontinuous. Discuss the continuity of the function f(x) on the closed interval. We have f(x) is equal to 2 over x^2+9. We are looking for the continuity on the closed interval of [-2,3]. So let's first see if we can check what is going on at the left side of our closed interval. Well
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the limit as x approaches -2, and it is the left side so it is going to be approaching from the right, of 2/(x^2-9). That is going to be equal to 2 over (-2)^2-9 which is going to be 2 over 4-9, or 2 over -5. Now as long as the limit of the function is equal to the function's actual value on this lower side of our closed interval, then it is going to be continuous at least on the left end. So we have f(-2). I was able to do direct substitution
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with this function to get the limit so I know that f(-2) is the same value. That is going to be -2/5 as well. Now let's check for the limit on the right side of the closed interval. Let's put this over here. The limit as x approaches 3, now this is the right side or the upper limit of our closed interval, so we are going to be approaching it from the left of 3, of 2 over x^2-9. We are going to have a problem here. Because 2/(3^2-9) is going to be 2 over
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9-9 which is equal to 2/0. That means that it does not exist. Well in the previous example I was able to manipulate the function by rationalizing the numerator to come up with another function that would agree. I was able to take that value, plug it in, and get the limit. Here I cannot do any fancy tricks. I can factor this to be (x+3) times (x-3). But there is no math or algebraic steps to manipulate this into another function that would allow me
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to plug in the value of 3. So the limit does not exist on the upper limit or domain of our closed interval. So this is not going to be continuous on the closed interval. Now let me just.... I did this analytically with some numbers. But you could also remember that this is a rational function and I said that rational functions are continuous for all values within their domain. I hinted to that over here. If I take my original function
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and say f(x) is equal to 2 over (x+3)(x-3), then the domain of this function is all of my x's going from negative infinity to negative three then picking back up between -3 and 3, and then after I have to skip here... See these zeros in the denominator have nothing to cancel with in the numerator, so they are going to be vertical asymptotes. I have to skip over that three, then the rest of my domain is 3 to infinity. 3 was part of my
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closed interval so I knew this was going to fail the continuity test before I even started. By the way the function basically looks like this. We have an x and y axis. If you are allowed to play with your graphing calculator.... There is very good chance your teacher is not allowing you to use your graphing calculator because it can do all the work for you allowing you to answer a question without really understanding what is going on. The function looks like
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this. Here we see that x value of 3, the upper limit of our closed interval, and the function clearly is undefined at that value. WHOOO!!! Next example. Almost next example. I just wanted to clarify something here. While we have failed the test for this being continuous on the closed interval, it would be continuous on the open interval from -2 to 3. Because it is just going to be continually going down to negative infinity. As long as I don't need
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it to be continuous at the value of 3 which is just one of the two places where this function has a restriction on the domain. That would be continuous on the open interval. But it again did fail the closed interval test for continuity. Find the values of x, if any exist, at which f(x) is not continuous. Ok. Well, I have a rational function. Like I said. Most functions like rational functions are going to be continuous everywhere where it is defined,
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or for all values in its domain. So I just have to find out what the limits are for this domain. I have been asking that question since Algebra 1 I think. At any rate, we are going to factor the denominator, if it can be factored. Otherwise I am going to need to solve for the denominator =0 using the quadratic formula. This is going to be equal to x-1... and factoring trinomials are very simple as the leading coefficient is equal to 1. It is a little
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bit more work if it is not. We just have to ask if there are factors of the constant that add or subtract to the middle term. For negative three it is going to be negative one and positive three. So it is going to be x-1 and x+3. Now, the factor of x-1 is going to create a restriction in the domain of x=1. Then we have x=-3. Actually for domain, this is not really x equals, but what x cannot equal. There is a hole in the graph at x=1 and there is a vertical asymptote
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at -3. The function looks something like this. There is a horizontal asymptote at y=0. It looks this. You can see the importance of what you learned in Algebra 2 or PreCalculus about graphing rational functions and finding those holes and vertical asymptotes. This function is continuous everywhere except for 1 and -3. Now the hole is a removable discontinuity. There is a removable discontinuity at x=1 and there is a non-removable discontinuity
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at x=-3. Over here piecewise functions again generally give us the most amount of work in finding possible places where it is discontinuous. This is just a constant function that is continuous in all of values of it's domain of all real numbers. This is a polynomial so that is continuous everywhere. This is another polynomial. So all three of these functions individually are continuous along all real numbers, negative infinity to positive infinity. So all we have
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to do is check to see if it connects at these values of negative one and positive one. These are very easy to graph so the best or easiest way to do this is to simply draw yourself a picture. You don't need a graphing calculator for this. We have a function of y=-2. That is a horizontal line going through the y value of -2. We are going to -1. So there is y=-2 and this is just simply less than -1, therefore we are going to put an open dot there which
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we may or not fill up. If we cannot fill it up, we have found a discontinuity there at negative one. We have 3x+1. Well y=3x+1 is a line with a slope of 3 and y intercept of 1. If I come up here to y=1, I can say there is my y intercept. The slope is 3/1, so up three and over 1. I can also do down three and over 1. Look at there. I am filling in the hole. So this goes from... Actually I was there a little premature there. That is
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x is greater than or equal to -1. If it was just greater than -1, I would still have an open dot. But it is equal to so I am going to fill up that space. I am going to, where x=1 and it is just x<1 so open dot. That is about as straight as I can get it. This is y=x^2, a parabola. let's see here. zero squared is zero. There is a point there but I don't really want to include that. I am really concerned about what the graph is starting to do at
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the x value of 1. Well 1 squared is 1. So we have go (1,1) from plugging in one and doing one squared is one. It is x is greater than or equal to 1. That is a closed dot and it is going to look something like this. So I can just see by drawing this that the left... the limit of the function as x approaches -1 from the left and approaches -1 from the right... they are both the same. The left and right limit are both equal to -3, so thus
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the two sided limit where x approaches negative one is equal to negative three. So this function is continuous at -1. But as I get to positive 1, I am approaching 1 from the left and my limit seems to be approaching 4. I can simply plug in 1. This is just a line, I can find the limit using direct substitution. I will write this down in a second. It is approaching 1, 2, 3, 4, and from the the right it is approaching a y value of 1. I can visually look at this
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and see that the function is going to be discontinuous at 1 because the left and right limits are not equal, thus the limit at 1 does not exist. That fails one of the continuity tests. I am going to pause and show you the algebraic work to find that left and right limit as well. Last example. Find the constant value of a that makes the function continuous. In this case we are going to make it continuous at zero. We have f(x) is equal to 2sin(x)/x
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and 2a... that is what the question is about, finding the value of a to make this function continuous at this value of 0 where the piecewise function is changing gears between these two functions.... -5x. So let's start with the part of the function where I am not actually find out what a is equal to so I have something to set this equal to and solve for a. So let's find the limit as x approaches zero. Now this is for when x is less than 0, so we are approaching
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from the left. We are going to find the limit as x approaches 0 from the left. Ok. We are going to be doing that for the first part of this piecewise function. I am just going to write 2sin(x)/x. Now to really make this a little bit clearer to how I am coming up this answer, I am going to rewrite this a little bit. I want to remind you that this is 2 and put a coefficient of 1 in front of the x. We are going to rewrite this as 2 times
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sin(x)/x. Because hopefully you are remembering that sin(x)/x, when you try to put in zero... I am dividing by zero. We talked about a special case of finding the limit as x approaches zero of sin(x)/x. You should have had that in one of your previous lessons or videos that I talked about in finding limits of trig functions. That is going to be 2 times 1 which is equal to 2. So the left limit, the left side limit at 0 is equal to 2. Now let's set
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up the right side limit and make sure that it does equal 2. The limit... I am going to bring this up here. The limit as x approaches 0 from the right of... it is of f(x).. but as I approach from the right of 0, that means that I am using x values that are greater than zero and I am using the second part of my piecewise function...That is 2a-5x. I am not just trying to find the right side limit. I want to make sure that this function is
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continuous. That starts off with making sure that the limit exists, and then you make sure that the function is the same value as the limit. Then it is going to be continuous. I am using direct substitution. So if I find the limit I know the function at that value is going to be the same. I want to make sure that this right hand limit is equal to 2. So the left and right hand limits are equal, thus the two sided limit exists. I am using
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direct substitution so I know the function is going to be the same value at 0, or c as your notes said. So that is what the right hand limit needs to be. I am letting x approach zero. There is two variables in here, but I am finding the limit as x approaches zero from the right. So x is approaching zero and I am going to do that substitution and say that 2a-5(0) is equal to 2. Well then that is just zero. We get 2a=2. Divide by 2 and
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that means that a is equal to 1. So if a=1, that is going to be the value that is going to allow this function... this piecewise to be continuous at 0 because our two sided limit and the function of f(x) at zero are going to be the same. The function is continuous. I am done. Thank you for watching. I know there was a lot in this video. But hopefully now you have a complete understanding of continuity on open and closed intervals. One sided limits....
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yada yada yada. I am Mr. Tarrou. BAM!!! Go Do Your Homework:)
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