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
eq a ) and is continuous at ( a ). (if an answer does not exist, enter none.)
(a) ( f(x)=\frac{x^{3}-1}{x - 1},a = 1)
the discontinuity is removable.
the discontinuity is not removable.
( g(x)=)
(b) ( f(x)=\frac{x^{3}-x^{2}-6x}{x - 3},a = 3)
the discontinuity is removable.
the discontinuity is not removable.
( g(x)=)
(c) ( f(x)=lfloorsin x
floor,a=pi ) (recall that ( lfloor h(x)
floor ) means the largest integer that is less than or equal to ( h(x) ).)
the discontinuity is removable.
the discontinuity is not removable.
( g(x)=)

Explanation:

Step1: Analyze function (a)

Simplify $f(x)=\frac{x^{2}-1}{x - 1}=\frac{(x - 1)(x + 1)}{x - 1}=x + 1$ for $x
eq1$. The discontinuity at $x = 1$ is removable. The new function $g(x)=x + 1$ is continuous at $x = 1$.

Step2: Analyze function (b)

Factor $f(x)=\frac{x^{3}-x^{2}-6x}{x - 3}=\frac{x(x^{2}-x - 6)}{x - 3}=\frac{x(x - 3)(x+2)}{x - 3}=x(x + 2)$ for $x
eq3$. The discontinuity at $x = 3$ is removable. The new function $g(x)=x(x + 2)$ is continuous at $x = 3$.

Step3: Analyze function (c)

The function $y=\lfloor\sin x
floor$ has a non - removable discontinuity at $x=\pi$ because the left - hand limit $\lim_{x\to\pi^{-}}\lfloor\sin x
floor = 0$ and the right - hand limit $\lim_{x\to\pi^{+}}\lfloor\sin x
floor=- 1$.

Answer:

(a) The discontinuity is removable. $g(x)=x + 1$
(b) The discontinuity is removable. $g(x)=x^{2}+2x$
(c) The discontinuity is not removable. $g(x)=$ONE