Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 43

Determine 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 $a$.

$\begin{equation} \begin{aligned} & \text{(a )} f(x) = \frac{x^4 - 1}{x - 1}, a =1 \\ & \text{(b )} f(x) = \frac{x^3 - x^2 - 2x}{x- 2}, a = 2 \\ & \text{(c )} f(x) = [[\sin x]], a = \pi \end{aligned} \end{equation}$




(a) $\displaystyle f(x) = \frac{x^4 - 1}{x - 1}, a =1$


$\begin{equation} \begin{aligned} \lim \limits_{x \to 1} \frac{x^4 - 1}{x - 1} =& \frac{(x^2 - 1) (x^2 + 1)}{x - 1} = \frac{(x + 1) \cancel{(x -1)} (x ^2 + 1)}{\cancel{x - 1}} = (x + 1)(x^3 + 1)\\ \\ =& x^3 + x^2 + x + 1 \end{aligned} \end{equation}$


The function $\displaystyle \frac{x^4 - 1}{x - 1}$ has a removable discontinuity at $a = 1$.
However, $g(x) = x^3 + x^2 + x + 1$ agrees with $f$ for $x \neq a$ and that $g(x)$ is continuous at $a = 1$

(b) $\displaystyle f(x) = \frac{x^3 - x^2 - 2x}{x- 2}, a = 2$


$\begin{equation} \begin{aligned} \lim \limits_{x \to 2} \frac{x^3 - x^2 - 2x}{x - 2} =& \frac{x(x^2 - x - 2)}{x - 2} = \frac{x [\cancel{(x - 2)}(x + 1)]}{\cancel{x - 2}} = x(x + 1)\\ \\ =& x^2 + x \end{aligned} \end{equation}$



The function $\displaystyle \frac{x^3 - x^2 - 2x}{x - 2}$ has a removable discontinuity at $a = 2$. However, $g(x) = x^2 + x$
agrees with $f(x)$ for $x \neq a$ and that $g(x)$ is continuous at $a=2$.

(c) $\displaystyle f(x) = [[\sin x]], a = \pi$



$\begin{equation} \begin{aligned} \lim \limits_{x \to \pi^-} [[\sin x]] =& \lim \limits_{x \to \pi^-} [[\sin x]] = \lim \limits_{x \to \pi^-} 0 = 0\\ \lim \limits_{x \to \pi^+} [[\sin x]] =& \lim \limits_{x \to \pi^+} [[\sin x]] = \lim \limits_{x \to \pi^+} -1 = -1 \end{aligned} \end{equation}$


The function $[[\sin x]]$ has a jump discontinuity instead of removable discontinuity. Therefore, the
$\lim\limits_{x \to \pi} [[\sin x]]$ does not exist