Question

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 ≠ a and (a) ( f(x) = \frac{x^4 - 1}{x - 1}, a = 1 ) ( \bigcirc ) the discontinuity is removable. ( \bigcirc ) the discontinuity is not removable. ( g(x) = square ) (b) ( f(x) = \frac{x^3 - x^2 - 56x}{x - 8}, a = 8 ) ( \bigcirc ) the discontinuity is removable. ( \bigcirc ) the discontinuity is not removable. ( g(x) = square ) (c) ( f(x) = llbracket sin(x)
rbracket, a = pi ) (recall that ( llbracket h(x)
rbracket ) means the largest integer that is less than or equal to ( h(x) ).) ( \bigcirc ) the discontinuity is removable. ( \bigcirc ) the discontinuity is not removable. ( g(x) = square )

Explanation:

Part (a)

Step1: Factor the numerator

We know that \(x^4 - 1\) is a difference of squares, which can be factored as \((x^2 + 1)(x^2 - 1)\). And \(x^2 - 1\) is also a difference of squares, so \(x^4 - 1=(x^2 + 1)(x + 1)(x - 1)\). So the function \(f(x)=\frac{x^4 - 1}{x - 1}=\frac{(x^2 + 1)(x + 1)(x - 1)}{x - 1}\) for \(x
eq1\).

Step2: Cancel the common factor

We can cancel out the common factor \((x - 1)\) for \(x
eq1\). So \(f(x)=x^3 + x^2 + x + 1\) for \(x
eq1\). Since the limit as \(x\to1\) of \(f(x)\) exists (we can directly substitute \(x = 1\) into \(x^3 + x^2 + x + 1\) to get \(1 + 1+1 + 1 = 4\)), the discontinuity at \(x = 1\) is removable. And the function \(g(x)\) that agrees with \(f(x)\) for \(x
eq1\) and is continuous at \(x = 1\) is \(g(x)=x^3 + x^2 + x + 1\) (or we can also write it as \(g(x)=x^3+x^2+x + 1\), which is the same as \((x^2 + 1)(x + 1)\) expanded).

Step1: Factor the numerator

We factor the numerator \(x^3 - x^2 - 56x\). First, factor out an \(x\): \(x(x^2 - x - 56)\). Then factor the quadratic \(x^2 - x - 56\). We need two numbers that multiply to \(- 56\) and add to \(-1\). The numbers are \(-8\) and \(7\). So \(x^2 - x - 56=(x - 8)(x + 7)\). So the numerator is \(x(x - 8)(x + 7)\). Then the function \(f(x)=\frac{x(x - 8)(x + 7)}{x - 8}\) for \(x
eq8\).

Step2: Cancel the common factor

Cancel out the common factor \((x - 8)\) for \(x
eq8\). We get \(f(x)=x(x + 7)=x^2+7x\) for \(x
eq8\). The limit as \(x\to8\) of \(f(x)\) exists (substitute \(x = 8\) into \(x^2+7x\): \(8^2+7\times8=64 + 56 = 120\)), so the discontinuity at \(x = 8\) is removable. And \(g(x)=x^2 + 7x\) (or \(g(x)=x(x + 7)\)) is the function that agrees with \(f(x)\) for \(x
eq8\) and is continuous at \(x = 8\).

Recall that \(\lfloor h(x)
floor\) is the floor function (greatest integer less than or equal to \(h(x)\)). We know that \(\sin(\pi)=0\). Let's consider the left - hand limit and the right - hand limit as \(x\to\pi\).

For \(x\) slightly less than \(\pi\) (i.e., \(x\in(\pi-\epsilon,\pi)\) for small \(\epsilon>0\)), \(\sin(x)>0\) and \(\sin(x)\) is close to \(0\). So \(\lfloor\sin(x)
floor = 0\) when \(x\) is slightly less than \(\pi\) (since \(0<\sin(x)<1\) for \(x\in(0,\pi)\) and \(x
eq\frac{\pi}{2}\)).

For \(x\) slightly greater than \(\pi\) (i.e., \(x\in(\pi,\pi + \epsilon)\) for small \(\epsilon>0\)), \(\sin(x)<0\) and \(\sin(x)\) is close to \(0\). So \(\lfloor\sin(x)
floor=- 1\) (since \(-1<\sin(x)<0\) for \(x\in(\pi,2\pi)\) and \(x
eq\frac{3\pi}{2}\)).

The left - hand limit \(\lim_{x\to\pi^{-}}\lfloor\sin(x)
floor = 0\) and the right - hand limit \(\lim_{x\to\pi^{+}}\lfloor\sin(x)
floor=-1\). Since the left - hand limit and the right - hand limit are not equal, the limit \(\lim_{x\to\pi}\lfloor\sin(x)
floor\) does not exist. A removable discontinuity requires that the limit exists (but the function is not defined or has a different value at the point). Since the limit does not exist, the discontinuity is not removable.

Answer:

The discontinuity is removable. \(g(x)=x^3 + x^2 + x + 1\) (or \(g(x)=(x^2 + 1)(x + 1)\) or \(g(x)=x^3+x^2+x + 1\))

Part (b)