How do you find the limit (x^3+4x+8)/(2x^3-2) as x->1^+?

Dec 27, 2017

$+ \infty$

Explanation:

use direct substitution:

${\lim}_{x \rightarrow {1}^{+}} \frac{{x}^{3} + 4 x + 8}{2 {x}^{3} - 2}$

$= \frac{{\left({1}^{+}\right)}^{3} + 4 \left({1}^{+}\right) + 8}{2 {\left({1}^{+}\right)}^{3} - 2}$
(${1}^{+}$ means a number slightly larger than 1, such as 1.0000001)

$= \frac{{1}^{+} + {4}^{+} + 8}{2 \left({1}^{+}\right) - 2}$

$= \frac{{13}^{+}}{{2}^{+} - 2}$

$= \frac{{13}^{+}}{{0}^{+}}$

$= + \infty$

graph{(x^3+4x+8)/(2x^3-2) [0.9, 1.1, -1000, 1000]}
from the graph, you can see that the function approaches $+ \infty$ as x approaches 1 from the right side