sketch-the-graph-of-the-function-and-use-it-to-determine-the-values-of-a-for-which-displaystyle-li-2

Question

Sketch the graph of the function and use it to determine the values of $ a $ for which $ \displaystyle \lim_{x	o a}f(x) $ exists.
$ f(x) = \left\{
          \begin{array}{ll}
              1 + \sin x  & \mbox{if $  x < 0 $}\
              \cos x    & \mbox{if $ 0 \le x \le \pi $}\ 
              \sin x  & \mbox{if $ x > \pi $}
           \end{array} ight.$

Anjali Kurse · Accepted Answer

Okay let&#x27;s get your graph of this piece wise function and use it to determine the values of A. For which the limit as X approaches. A. For F. A bex exists. So let&#x27;s just let&#x27;s just grab this so we&#x27;ll draw a nice long coordinate plane here. Okay let&#x27;s start at the left most side. Let&#x27;s start with the one plus sign X. for X is less than zero. So all I&#x27;m going to do here is right down the pieces that I want to draw properly. Right the X. And the Y. Value. So we can have a nice pretty graph. So let&#x27;s start here. The sign of -2 Pi is just zero but the sign is transposed up one because of the one plus in front of that. So all we&#x27;re going to do is move the sine function up one. Now the sign of negative three pi halves is positive one, add one to that. Make it too. And then we just draw r sine function from there And now this is important. Am I going to have an open circle or a closed circle at one? I am going to have an open circle because it&#x27;s X. Is less than zero. Not X is less than or equal to. Now let&#x27;s do the second half. This is going to be co sign from zero two pi and then from there on um sign okay for the co sign peace. The coastline of zero is one. So we are going to end up filling in the circle just because it&#x27;s equal to one. So the function does continue their. The coastline of pie halves. Zero. The cosine of pi is negative one of each other there Sign up I is negative one. Now is this a closer an open circle? It&#x27;s a close circle. Woo hoo. Okay now the third part this is going to be the side X from X is greater than pot. So where does sign what a sign of pot equal to? It&#x27;s equal to zero. This is really important here. Sign of three Pi Halves Is equal to -1 and sign of two pi equal to zero. So it&#x27;s just going to continue like this. So not for the second part of the question which is determining the values of a for which the limit exists. So there are two key places to to think about here. That&#x27;s going to be the transition points. The transition between functions in the piecewise. So that&#x27;s zero and that&#x27;s pie. Let&#x27;s talk about 0 1st zero. So from the left hand side the limit as X approaches zero. The function we&#x27;re thinking about here is one plus Synnex. What is that equal to sine of zero is zero plus one is one. Now Still in the zero. Let&#x27;s talk about the coastline. Oh I should change the color to green here and that&#x27;s the color. I did my co sign the limit As X approaches zero from the right because remember coastline it&#x27;s the right hand side of this part of Cosine X Is also equal to one. So because the limit from the left matches the limit from the right, we know this is the limit does exist. The limit as X approaches zero of F of X exists. You can also see that just from the graph itself, see how it continues. There aren&#x27;t any sudden stops. There&#x27;s no holes is perfectly continuous. Now the other problem point is going to be Pie that we need to evaluate. So let&#x27;s do pie. Let&#x27;s do it from the left first pi minus. And that&#x27;s going to refer it to the co sign. Co sign of Pie is zero. It&#x27;s negative one. Sorry, negative one. Totally know what I&#x27;m doing. Okay. Hi, again. Okay. The limit as X approaches pi plus of this is the last function here. It&#x27;s after pie. So we&#x27;re gonna be talking about sine is equal to zero. The limit from the left does not match the limit from the right. It doesn&#x27;t therefore, which is what these three dots mean? The limit as X approaches. Oh, I&#x27;m sorry, I just realized that this is zero not part that should say Pie. Hi, selects the limit as X approaches pi of F of X. Doesn&#x27;t exist. Okay, so remember with this piece wise, the first step is to graph and then the second step to determine if the limits exist. You have to look at the problem points which is where there&#x27;s gonna be a transition between these functions and that&#x27;s where a problem could occur. That&#x27;s how we check if it&#x27;s continuous, if the limit is actually there, and in this case it exists as X approaches zero, but does not exist as X approaches pi.

Leon Druch · Answer

defined in three different parts. Uh function is defined to be one plus X. When x is less than negative one, that&#x27;s the red portion of the graph that you see. The function f of X equals X squared. When uh negative one is less than or equal to ex uh ex in turn is strictly less than one. That&#x27;s the blue portion of the graph that you see. And last but not least when x is greater than or equal to one. Uh F of X is defined to be two minus X. We want to find all values a along the X axis for which the limit exists. And this graph is uh going to greatly help us. You can see um that this uh function is basically connected. There&#x27;s no breaks anywhere in the function except right here. When X is negative one, we have a break into function when X is positive one. Uh You can see the blue portion of the function uh meets nicely uh the green portion of the function. And so the limit as X approaches one from the negative side, the blue side is going to be the same limit as X approaches one from the uh positive side. The green side. But over here, it when uh X is negative one, or specifically when a is equal to negative one. Uh This function uh does not have a limit in other words. Uh The limit of f of X as X approaches negative one does not exist. Um The limits exist for dysfunction for any other value, any other real number value along the X axis, it does not exist. The limit of dysfunction does not exist when X approaches the value of negative one because as X approaches negative one from the negative side to function uh approaches zero. So the limit of F of X as X approaches negative one from the negative side is zero, but the limit of F of X pay attention to the blue portion of the curve. Now the limit of F of X as X approaches negative one from the positive side, uh F of X is approaching the value of one. And so the limit of F of X as X approaches negative one does not exist because the limit of F of X as X approaches negative one from the negative side does not equal the limit of F. Of X as X approaches negative one from the positive side. And you can clearly see that because the red portion of the graph is getting close to zero, think of the Y axis as to function values, whereas the blue portion of the graph is getting close to the value of one

Bon Zapata · Answer

So we need to make a graph of this function and use it defined where the limits exist. So I&#x27;m gonna be doing this and decimals feel free to use any sort of graphing calculator. The only tricky step might be getting these restraints up. I&#x27;ll explain how to do it in decimals, but each program is gonna have its own way of doing active. Okay, so the 1st 1 I&#x27;m just going Teoh, assign this a function name. So after backs is equal to one plus sign racks and to do the restraint, I put a curly bracket from then X is less than zero for the next one. So let me go back for a minute. Even though this is all one function just to not confuse the graphing calculator, I&#x27;m gonna make this free separate functions of. So this 1st 1 was f of X. Then the next one will be tree of axe and then a tree, Becks. So let&#x27;s go back. She of axe is equal to co sign of X arm. Strength is that zero is less than or equal to X is less than or equal to pi. And then our last one will be a true backs and this is equal to sign of acts. Assuming that X is greater than I both from looking at this, we need to figure out where the graph is continuous. The one point of interest is right here will receive this. A jump from the second graph to the feared graph up. You can see we cannot draw this entire graph without lifting up the pencil from the page. We have to lift it up right here. So this gives us a dis continuity and the limit does not exist at this one point. At what point is this? It&#x27;s the point where we switch from graft to to graft free and we switch at pie. That&#x27;s the last point of the 2nd 1 and the first point of the fared line. So this graph is continuous, which means that the limit exists everywhere except at high

James Kiss · Answer

we&#x27;ll start with the graphing calculator using justice Graphing calculator online. We have the function at the X equals X times, the absolute value she opened. Keep and you&#x27;ll see the absolute value o X and so we can use the grid lines as our horizontal line for the test and nowhere. Even if we zoom in, is it going to intersect the graph more than once? What? CASS is the horizontal line test, meaning that this function is a 1 to 1 function.

Ankit Gupta · Answer

the given function is ffx is equal group X games absolute value of X. Now the first thing we&#x27;ll do is blood dysfunction on a graphing calculator. So this image here shows the graph off this function Informix as given by a graphing calculator. No, What we can see here is that there is no such horizontal straight language and draw which will intersect this car or this graph at two different points. Which means that we can see for this particular craft f of X one is not equal toe ever x two if x one is not equal to X to this is for any two different values off Exxon A next to for this girl. Therefore, if off X is I want to one function.

Heather Zimmers · Answer

all right, we&#x27;re going to graft this piece wise function. So let&#x27;s start by thinking about what each piece is expected to look like. So we have f of X equals the absolute value of X. And if you think about that function, it typically looks like a V shaped graph with its Vertex at 00 And then we have f of X equals one. So if you think about what Y equals one looks like it would look like a horizontal line at a height of one. So we&#x27;re going to have a piece of each of those. And the second part of the statement tells us the domain for each piece. When you see the absolute value of X is less than or equal to one, it means the same thing as negative. One is less than or equal to X is less than or equal to one. And when you see that the absolute value of X is greater than one, it means X is greater than one or X is less than negative one. Okay, so that means we&#x27;re going to have the absolute value graph, but only the part that falls between negative one and one. So that&#x27;s going to be from this point to this point. We have that V shaped graph. Beyond that, we&#x27;re going to have the horizontal line, so we&#x27;re going to have a horizontal line at a height of one to the right and to the left of that be and there&#x27;s a graph.

Anjali Kurse · Answer

This is problem number twelve of Stuart Calculus. Eighth edition. Section two point two gets the graph of the function and use it to determine the values of a fur. What&#x27;s the limit and sex approach? Izzy of the function Athletics exists, the function of select is shown. Here is a piece for dysfunction, and there are three Pacific functions that we&#x27;re going to plant for their respective intervals. Want the Sonics is the first one. Cose, an imex is the second. Santa picks is the third, so we&#x27;re going first graphed dysfunction notice. The first interval is lesson zero. So to the left of the X axis are sorry to the left of the Y axis, and we&#x27;re going to plug in these points. They see here parts of pie because since signing co saner trigonometry functions, it will make our calculation that much easier. So if we choose that negative pie is the first X value to plot using the calculator sign of negative pipe gives us about ur zero. One plus hero will be one that&#x27;s the first point were implied the next point where menus and plot is this half a pie. Negative power, too. Sign of negative power over to using a calculator is negative one one plus think of the one O. P. Zero and presently at X equals zero. We know that dysfunction Chinas or one plus sign of zero we&#x27;LL approach one plus zero or one and the sine function as a distinct Kirby Look to it and it will have a whole hero zero because this&#x27;s not less than are equal to it is only equal to. And over here the function will look a little like that. This is what the sign function looks like. And the next interval from zero to pine notice that thie interval includes a zero and pi. So we&#x27;re going to d&#x27;Oh. We&#x27;re gonna plant coursing from zero to pine, beginning with zero using your calculated cosign. A zero is one. So we see that these two or the point is defined in the hole from the previous graph. If we continue into pirate too constant a pirate too. We know his hero from a calculator and then causing a pie. The last point is thinking of one. I want to be sure toe that we know. Ah, there. We recall that like the same function because in function has this distinct shape. Finally, for the last interval Sign of X will be planted when X is greater than pie. But first we must know where to start. So we&#x27;re going to use X equal to pi. I&#x27;m making sure that we dropped a hole because it&#x27;s not equal to pi. And then we use a calculator sign of pie. We know to be zero. I mean that there&#x27;s a hole right there and then we continue graphing for three prior to All right, that here three hi or to using a calculator kids us negative one. And then at two pi, we get sign of two by zero again. So our sand function will look like that. Continue on after to be completed the graph of athletics, which is made up of three different functions. And we finally answer her question for which kinds of AIDS does this limit? Ah, there&#x27;s a limit as X purchasing of the function have exist. As long as thie function is continuous, the limit will exist at each of the points here. At each of the X equals eight points atnegative pine. If a is equal to negative point. We know the limit exists because it&#x27;s the same from the left in the rain and even at a point like this approaching zero from left, the limit is one approaching zero from the right. The limit is one. Therefore, element exists at this point, the issue or the our concern is when the function has a job. So from here, the limit from the left of F as it approaches ty, the limit is negative one. But as it we follow the function from the right towards the pie, we see that the limit is zero. And since these both do not agree, we say that the element when hey is equal to hi, the limit is un exists. So our final answer that the limit exists for all values may accept a spray.

Anjali Kurse · Answer

Okay here we have a photograph of the piecewise defined function F of X one plus one plus sine of X. When X is less than zero. F of X is defined to be co sign of X for X between zero and pi inclusive, and F of X equals sine of X. When X is greater than high. Looking at the graph of F of X, we can see that it is continuous everywhere, except at X equals pi. Uh, So the limit of F of X as X approaches some number A will exist for every number A except high.

Problem

Use the graph of the function $ f $ to state the …

Problem 12 Medium Difficulty

Sketch the graph of the function and use it to determine the values of $ a $ for which $ \displaystyle \lim_{x\to a}f(x) $ exists.
$ f(x) = \left\{
\begin{array}{ll}
1 + \sin x & \mbox{if $ x < 0 $}\\
\cos x & \mbox{if $ 0 \le x \le \pi $}\\
\sin x & \mbox{if $ x > \pi $}
\end{array} \right.$

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