Answer
a. $g(x) = (x^{2} + 1)(x+1)$; The discontinuity can be removed by defining $f(1) = 4$.
b. $g(x) = x^{2} + x$; There is a removable discontinuity at $x = 2$.
c. The discontinuity at $x = \pi$ is a jump discontinuity so $\lim\limits_{x \to \pi} f(x)$ doesn't exist.
Work Step by Step
a. $f(x) = \frac{x^{4} - 1}{x-1}$
Expand the function:
$f(x) = \frac{(x^{2} + 1)(x^{2}-1)}{x-1}$
$f(x) = \frac{(x^{2} + 1)(x-1)(x+1)}{x-1}$
Cancel out $(x-1)$:
$f(x) = (x^{2} + 1)(x+1)$
Now we find the limit using the number $a$ given by the problem.
$\lim\limits_{x \to 1} (x^{2} + 1)(x+1) =$
$\lim\limits_{x \to 1} (1^{2} + 1)(1+1) =$
$ (1 +1) (2) = 4$
b. $f(x) = \frac{x^{3} - x^{2} -2x}{x-2}$
Factor $x$:
$f(x) = \frac{x(x^{2} - x -2)}{x-2}$
Expand:
$f(x) = \frac{x(x+1)(x-2)}{x-2}$
Cancel out $(x-2)$:
$f(x) = x(x + 1)$
Now we find the limit using the number $a$ given by the problem.
$\lim\limits_{x \to 2} x(x+1) =$
$\lim\limits_{x \to 2} 2(2+1) =$
$ 2(3) = 6$
c. $f(x) = sinx$
$\lim\limits_{x \to \pi^{-}} sinx =$
$\lim\limits_{x \to \pi^{-}} 0 = 0$
$\lim\limits_{x \to \pi^{+}} sinx =$
$\lim\limits_{x \to \pi^{+}} -1 = -1$
Since both limits have different results we know now that the limit doesn't exist.
