Problem

Let $ g(x) = \frac{x^2 + x - 6}{| x - 2 |} $. …

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Problem 48 Hard Difficulty

Let $ g(x) = \text{sgn}(\sin x) $.

(a) Find each of the following limits or explain why it does not exist.
(i) $ \displaystyle \lim_{x \to 0^+}g(x) $
(ii) $ \displaystyle \lim_{x \to 0^-}g(x) $
(iii) $ \displaystyle \lim_{x \to 0}g(x) $
(iv) $ \displaystyle \lim_{x \to \pi^+}g(x) $
(v) $ \displaystyle \lim_{x \to \pi^-}g(x) $
(vi) $ \displaystyle \lim_{x \to \pi}g(x) $

(b) For which values of $ a $ does $ \displaystyle \lim_{x \to a}g(x) $ not exist?
(c) Sketch a graph of $ g $.

Answer

a. $\begin{array}{llllll}{\text { (i) } 1} & {\text { (ii) }-1} & {\text { (iii) DNE }} & {\text { (iv) }-1} & {\text { (v) } 1} & {\text { (vi) DNE }}\end{array}$
b. $\lim _{x \rightarrow a} \operatorname{sgn}(\sin x)$ will not exist if $a=n \pi$
c. Answer not available

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Video Transcript

let G of X equal sign of sine of X. That is the sign of the sine function of X. You pray? We find each of the following limits or explain why it does not exist limit when X goes to zero from the right of G. Fx. The second one will be limit when x goes to zero from the left of G of X limit. When X goes to zero of the sum function limit. When eggs goes to buy from the right by from the left and by and in purview for which values of a does limit when X goes to a of the fx not exist. That is we a study where or in which point we have no limit any party with Katie graph of G. So we're starting with parsi. And remember first that function the sine function of a number Is equal to one. Is that number is positive, Is negative, one is the number is negative And zero gives them a very zero. So these functions will define a real real numbers because any real number going to be either positive, either negative or no. So this is a sine function and when we combine or compose that with the sine function a true parametric function sine of X. So we get so some sort of step function function that is Function that takes into devil is 1810 at all points of segments. Let's see what I'm talking about. So we're gonna sketch the function here. Let's try to write so we have this Okay, yeah, vertical axis, the Y axis. Let's say we have sine function at 00. So we had something like this. After that we had something like this, it repeat itself periodically like that. Let's see okay, here we have the same and this and this and that goes on. So here we have -1 Because he's a minimum value of sign And the maximum value is one. Now I'm going to try to draw the function of sign. That is our function assists side of sine of X. That is we apply the same function and after that we apply the Function SGN. So we applied that function is yet to all the eggs in this interval here from zero to buy. For example what begins is that all points there has emerged one because the sign of those points is positive at zero. The sinus zero. So the the image of zero is zero. The composition of two function cuba zero. First a sign of serious era and this sign at the end of zero is 0. So we have this point here after that all points from here do here has him image one because sign of those points is positive. And when we a place you into that we obtain one. And I'm drawing this open circles meaning That particular point for example by the images zero because we have signed the sine function At by a zero & SGN of Syria zero. What happened to the points from pie too to buy. Well, they're designed function is negative. Then all points there As Image 91. And we got to open the endpoints because those points in particular has image zero. So that repeat itself all the time. So we get this open here at this point the images serum. And then we have this playing here open here and so on. Let's do this. Oh there part of the graph of the composition. It is yet to be tangent to the coastline function withdrawal correctly. And here we have the same with this is open here here we have a zero. Mhm. We have the segment open here and here we have zero. We have zero here also. And we have open value turned in here and open here. That is the blue segments. And these separated points are the graph the sketch of the graph of G. So as we can see there are discontinuities in a bunch of funds and we're going to talk about that now when we talking about part A that is for example limited limit. When egg X goes to zero from the right of these functions. See blue throughout here, if we go to zero from the right that is we are coming this way to zero. We see that the image Of the composition is always one. And When we get closer and closer to see what we keep the value one all the time. It doesn't matter what happened to because the limit does not take that into account. We are only interested in what happened near Syria but not a serious at all. So the limit When X goes to zero from the right of she of X is one Because from the right, if she for instance is constantly equal to one when we are close to see her from the right now we do. Second part is the limit. When X goes from the left to zero, that is it comes to zero coming from this this way and all the values near here from the left, the all the balls of the functions are equal to negative one. So this limit is negative one. Mhm. And now the 3rd limits, there is the villa lateral limit, their bilateral limits at zero, there is the usual limited zero. That means that It doesn't take into account that we come from one side or another, but we are close to zero does not exist. And that is because to exist To the two lateral limits must exist and be equal. So it doesn't exist because the limit from the left and from the right are not equal. Wow, that's the reason that is, is a jump of the function is what we call the charm discontinuity of the function. Yeah, okay. At pi similar to CR four for former member is the limit when X goes to buy from the right. Yeah. And that if we see the graphs here from the right of by coming from here, function is always equal to negative one. So he's negative one. And the limit when we come goes to X goes to buy from the left, you would see the graph again coming to buy from the last the images are all equal to one. So the limit from the left by is what And again because these two lateral limits are different. The limit up by the bilateral limit that fast and that doesn't exist his ex. And we can see that these is uh the situation at all points where the functions changes sign or equivalently where the function Is equal to zero. So part B. Here we sketch a graph passes this part A And now I'm doing for B. Where are the values of A. Were the limit when eggs goes to a does not exist. Well, all the points for the function changes sign or equivalently in this case where the function is zero. So we know that sign is zero at the multiple the integer multiples of K. Of Peyser. So escape. I for K uh an integer number For example, that by as we saw at zero to buy three by I got to buy -2 piles or mhm And these are the only point for the function this is continuous. There are infinitely many of those but They are the only one where the functions is continues in other and all other points. The function is continuous. Yeah. So that is at the point where sign of eggs changes side. Mhm. Which in this case is equivalent to saying that we're sign of existing katsura. Remember that? It's not true for all functions, but because um if the function is continuous and change the sign, there's got to be serious. That's that's the theory. And we know. But the function can be tangent to have zero and be tangent to the axis. For example, X squared at zero. So in this case the continually as a function in place, that is the same As we're function is zero. So you say that also when we're science in design for or it is the same where sign of eggs is equal 20. And those are the points A of the form K. Time spy where K. Is an integral number. Yeah.

Oswaldo J.

Universidad Central de Venezuela

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Problem

Let $ g(x) = \frac{x^2 + x - 6}{| x - 2 |} $. …

Problem 48 Hard Difficulty

Answer

a. $\begin{array}{llllll}{\text { (i) } 1} & {\text { (ii) }-1} & {\text { (iii) DNE }} & {\text { (iv) }-1} & {\text { (v) } 1} & {\text { (vi) DNE }}\end{array}$
b. $\lim _{x \rightarrow a} \operatorname{sgn}(\sin x)$ will not exist if $a=n \pi$
c. Answer not available

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a. $\begin{array}{llllll}{\text { (i) } 1} & {\text { (ii) }-1} & {\text { (iii) DNE }} & {\text { (iv) }-1} & {\text { (v) } 1} & {\text { (vi) DNE }}\end{array}$b. $\lim _{x \rightarrow a} \operatorname{sgn}(\sin x)$ will not exist if $a=n \pi$c. Answer not available

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a. $\begin{array}{llllll}{\text { (i) } 1} & {\text { (ii) }-1} & {\text { (iii) DNE }} & {\text { (iv) }-1} & {\text { (v) } 1} & {\text { (vi) DNE }}\end{array}$
b. $\lim _{x \rightarrow a} \operatorname{sgn}(\sin x)$ will not exist if $a=n \pi$
c. Answer not available