a) Since both limits are distinct and do not exist, we conclude that x = - 1 is not part of the domain of the rational function.
b) The function [tex]f(x) = \frac{x}{x^{2}+ x}[/tex] is equivalent to the function [tex]g(x) = \frac{1}{x + 1}[/tex].
How to determine whether a limit exists or not
According to theory of limits, a function f(x) exists for x = a if and only if [tex]\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)[/tex]. This criterion is commonly used to prove continuity of functions.
Rational functions are not continuous for all value of x, as there are x-values that make denominator equal to 0. Based on the figure given below, we have the following lateral limits:
[tex]\lim_{x \to -1^{-}} \frac{x}{x^{2}+x} = - \infty[/tex]
[tex]\lim_{x \to -1^{+}} \frac{x}{x^{2}+x} = + \infty[/tex]
Since both limits are distinct and do not exist, we conclude that x = - 1 is not part of the domain of the rational function.
In addition, we can simplify the function by algebra properties:
[tex]\frac{x}{x^{2}+ x} = \frac{x}{x\cdot (x + 1)} = \frac{1}{x + 1}[/tex]
[tex]g(x) = \frac{1}{x + 1}[/tex]
The function [tex]f(x) = \frac{x}{x^{2}+ x}[/tex] is equivalent to the function [tex]g(x) = \frac{1}{x + 1}[/tex].
To learn more on lateral limits: https://brainly.com/question/21783151
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