## Calculus: Early Transcendentals 8th Edition

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.
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.