If $f(x)=left[frac{1}{sqrt{2}}(cos x+sin x)ight], 0<x<2 pi$, where $[.]$denotes the greatest integer function, then the number of points of discontinuity of $...

Which of the following functions $ f $ has a removable discontinuity at $ a $? If the discontinuity is removable, find a function $ g $ that agrees with $ f $ for $ x \neq a $ and is continuous at $ a $.

(a) $ f(x) = \dfrac{x^4 -1}{x - 1}$, $ a = 1 $
(b) $ f(x) = \dfrac{x^3 - x^2 - 2x}{x - 2} $, $ a = 2 $
(c) $ f(x) = [ \sin x ] $, $ a = \pi $

Yet square. So when you read here, so we have f of X is equal to X over square it of one minus co sign two x When we draw this it looked something like this and you see that there is a jump discontinuity at zero part B. We're gonna be focusing on the bottom part. We're gonna use the trick identity for a double angle co sign only, get one minus co sign of two X is equal to one minus one minus two. Signed square, which is equal to two signs square. When we calculate the dirt, the limit as X goes to the positive side of zero for X over square of one minus co. Sign two acts. This becomes equal to the limit as X approaches zero positive for X over square root of to sign square root of to sign Sign of X, which is equal to one over square root of two limit as X approaches. Zero positive of X. Over sign of X. This is equal to one over square root of two. This is from the right side, so it's gonna be positive. And when we find the derivative the limit. Excuse me. Um, since it's gonna be from the left side, it's gonna be negative.

No we have a question in which function is being given as one by x minus Flynn What X less than one and X q -2 x plus five. But access okay then if you want I'm going to find each point of discontinuity So there is only one critical point or we should say there is only one point which seems to be the and we see in twitter points of discontinuity which is X equal to Move one. Okay, so let us take the continue these contributors that x equal to one left hand continue to this left and limit will be limit when X approaches to one from left side affects okay from left side effects so this is want to buy one minus one because this will be the this will be there, what we say, this will be the function so this will approach to infinity right and limit, limit when X approaches to one from but I have said fx so this will come here so one q -2 into one plus 5 so forth we can see that left and limit is not right and limit. So it discontinues said Discontinues had x equal to one. Yeah there is only one point of this, continue to over here remember now and we will this is basically we can see that this is actually equal to one is a symptom of this, so if it is a symptom so it will be non removable discontinuity. Non removable. This continuity as it is an Essene tote thank you

So look at the function. You see that the man X minus one must be a critic. Could Juzo express echo to one? And for those will you hand the function is continuous going to Yes. So I hear the question. Ask us which of the following point is not a point under day after discontinuity. So the answer has to be there, eh x It could you one will be the point of the continuous

So for this question and seeing which of the following is true, But the graph of F X equals excellent for about five exits. Zero little choice. So it's either a corner, a cost of vertical tension or dis continuity, or it doesn't exist if there's no exist. So we've already dealt with problems that half of X equals, except for fifth. So if we remember from those with the answer is, then we're good to go. But to explain it one way that we can do this. There's Coolidge Window, where we can first graph it if we graph it. We see we have something that looks like this. We can go ahead and say, Yup, that's a cusp. But what else we can do is we can take the limit of ex approaching zero from the right side versus a limit of ex approaching zero from the left side of how we would take a derivative. So we have X minus half of a, which is looking zero in at zero over X minus a so ex same thing here B f of X over X, and this would give us one over X to the one over five. If we just get it out the right side, this is going to give us a positive 30. If you go from the left side, this is going to give us negative infinity. So we get different things when you're approaching from different sides. So that means that when we have a of expose except for fifth is not defensible at X equals zero. Because there we have a custom that I do it.