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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.

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So this problem deals with removable discontinuity. So if you were to plug in the value of one into this function, you would have an undefined denominator. So what we do is we factor in order to see if this discontinuity can be removed. So extra the fourth minus one can be factored into X squared minus one and X squared plus one. This is the difference of two squares where a. Is equal to X squared. Now I can keep going, here's another difference of two squares. This becomes X plus one, X minus one. The x minus ones are removed. So this is removable. And then my new function that I would use that would behave like the original except with a discontinuity would be this multiplied together X cubed plus X squared plus X plus one. Now for the next one, if you plug into you get something undefined in the denominator. So again it needs to be factored. So I start by factoring out the G. C. F, which is X. And then I'm going to factor, I'm going to factor that remaining part that quadratic. And I ask myself what are the factors of negative to that will add to give me negative one. And that would be X minus two X plus one. And they are the X minus two will cancel out. So my new function then that I could use that would behave like the original with a whole or a discontinuity is X squared plus X. And then finally foresee if I'm reading the notation correctly, this is a type of a step function. So when you're dealing with this type of step function you basically have two values for sign. You basically have zero and and negative one because in a step function you're you're rounding your values down So let me go to a different screen real quick and just show you I use graphing technology to help with this one. So this is the online program DEZ most dot com and I graft I grafted as a floor function. It doesn't really give me a step function availability. But you can see we have this little jump here. So at pi and this is not removable. I can't factor this at all so there's nothing I can do about this. Dis continuity here. Oops wrong screen. There we go. I'm sorry about that. Um So this is non removable.

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