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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 $
(a)$=\left(x^{2}+1\right)(x+1) \quad\left[\text { or } x^{3}+x^{2}+x+1\right]$ for $x \neq 1 .$ The discontinuity is removable and $g(x)=x^{3}+x^{2}+x+1$ agrees with $f$ for $x \neq 1$ and is continuous on $\mathbb{R}$(b) $=x(x+1) \quad\left[\text { or } x^{2}+x\right]$ for $x \neq 2 .$ The discontinuity is removable and $g(x)=x^{2}+x$ agrees with $f$ for $x \neq 2$ and is continuous on $\mathbb{R}$.(c) \[\lim _{x \rightarrow \pi^{-}} f(x)=\lim _{x \rightarrow \pi^{-}}[\sin x]=\lim _{x \rightarrow \pi^{-}} 0=0 \text { and } \lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{+}}[\sin x]=\lim _{x \rightarrow \pi^{+}}(-1)=-1, \text {so } \lim _{x \rightarrow \pi} f(x) \text { does not }\]exist. The discontinuity at $x=\pi$ is a jump discontinuity.
Calculus 1 / AB
Chapter 2
Limits and Derivatives
Section 5
Continuity
Limits
Derivatives
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This is problem number forty nine of the Stuart Calculus eighth Edition Section two point five. Which of the following functions F has a removable disk? Continuity at A. If the disk continuity is removed, find a function G that agrees with F for X is not equal to King and his continuous Eddie for party. Our function have his equal to X to the fourth minus one, divided by the quantity X minus one and is equal to one and ever for uh, f f removed removal of this continent thing First, it needs to have a discontinuity within the function itself. We see the function and that a nominator has it is continuity. Definitely X equals to one because that will make division zero possible, which is something that it's not allowed it for. A sequel to one. Ah, however, the only way to check if it is removable is to take the limit as X approaches a or one and determine whether Islam it exists For movable discontinuity, the limit must exist. Who we take the function next to the fourth minutes one over the quantity expense one. Ah, we factor the numerator x squared plus one X squared minus one, and we affect their once more as well. We did. We did this factory. And because it's a difference of squares X squared Ah, minus one squared. This can be simplified a little further hands ex scripts. One will deploy it by X plus one times X minus one divided by experience one, the experience one will cancel our limit. We're not ah, evaluating for X equal to one. We're approaching X equals one and this limit ends up being one squared plus one. That's two. Not a lot of buying one plus one wishes to or two times two is four. So parliament does exist. It's equal to four, all right. And this means that the dis continuity at X equals to one is actually removable. This community and because it's kind of these removal, we can name a function G equal to and we're in the sequel to is actually dysfunction here. That results from simplifying the function holdem simple. Fine. Um, essentially, what is left over? Expert plus one multiplied by experts. One is the function G. That is continuous city now, and that is our answer for party. Uh, for part B we take a similar approach particular that function a function for F is X cubed Maya's X squared of minus two x tomorrow by the quantity X minus two at the value A equals two and we see that too. Would it cause it is continuity in the denominator for this function. So again we use are limited to check. If this continuity are this discontinuities removable women's expertise to and the numerator, we can factor out one X and then we should be able to also factor this quadratic here accident my ex minus two times X plus one to write of my expense too. We can see here that this it will cancel this in the limit leaving us with X times X plus one devaluated X equals two gives us two times to piss warm or to temper three, which is six. So you know exists. And we know that the discontinuity at X equals two is thus removable and similar to the party Chief of X Is this leftover function here x more complaint? Max was home and that is our answer for the function cheat that agrees with us at all points X article too hey, but is definitely continuous at a So this function is continuous too. And that now you eyes defined my uncle six X equals two Finally, for part. See, we have the function f of X equals the greatest entered your function of sign of sine x. This raider interject for entered reflection is also known as the floor function. And we're looking at the condition that ese equal to pi narrow. We quickly turn a sketch for the function sign just to recall what this may look like and as the shape and prize right here. But this is the sine function, the floor function for sign. Well, actually be try blue. It'LL be a thicket. Take a look at To the left of pie is how he's running from zero two one. And so the function here its function or look like ah, zero up until pine and the negative one Great afterward. Take a left. The point it is zero because the value for assignment is between your own one. And if and if the number X is between zero and one, the floor function gives me the value of zero. Because that's the greatest center Jer I'm not greater than the value. And then for these negative values between zero negative one. Sigrid, isn't ginger not greater than the value itself? Ah is equal to negative one. So what we see here is that a piney this floor function of sign of X as a jumper discontinuity that time time it is most definitely not a removable. This Continuity Limited's expertise pine The left is not equal delimited and sex purchase by from right he left limit is equal to zero. The right Lim is equal to make it one. This is a chump just continuity not a removable. This continuity Ah! And so parts of the answer is that the function floor of San ethics does not have a removable this continuity.
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