## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 1^-} \frac{1}{x^3-1} = -\infty$ $\lim\limits_{x \to 1^+} \frac{1}{x^3-1} = \infty$
(a) $f(x) = \frac{1}{x^3-1}$ $f(0.9) = \frac{1}{(0.9)^3-1} = -3.69$ $f(0.99) = \frac{1}{(0.99)^3-1} = -33.67$ $f(0.999) = \frac{1}{(0.999)^3-1} = -333.67$ $f(0.9999) = \frac{1}{(0.9999)^3-1} = -3333.67$ $\lim\limits_{x \to 1^-}\frac{1}{x^3-1} = -\infty$ $f(1.1) = \frac{1}{(1.1)^3-1} = 3.02$ $f(1.01) = \frac{1}{(1.01)^3-1} = 33.0$ $f(1.001) = \frac{1}{(1.001)^3-1} = 333.0$ $f(1.0001) = \frac{1}{(1.0001)^3-1} = 3333.0$ $\lim\limits_{x \to 1^+}\frac{1}{x^3-1} = \infty$ (b) $f(x) = \frac{1}{x^3-1}$ If $x$ is slightly less than $1$ then $x^3$ is slightly less than $1$. Then $x^3-1$ is a negative number very close to zero. Then $\frac{1}{x^3-1}$ is a negative number with a very large magnitude. Thus $\lim\limits_{x \to 1^-}\frac{1}{x^3-1} = -\infty$ If $x$ is slightly more than $1$ then $x^3$ is slightly more than $1$. Then $x^3-1$ is a positive number very close to zero. Then $\frac{1}{x^3-1}$ is a very large positive number. Thus $\lim\limits_{x \to 1^+}\frac{1}{x^3-1} = \infty$ (c) On the graph, we can see that $~~\lim\limits_{x \to 1^-}\frac{1}{x^3-1} = -\infty~~$ and $~~\lim\limits_{x \to 1^+}\frac{1}{x^3-1} = \infty~~$