Question

F(x)

Answer 1

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 $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$.

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 $\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$. 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:

$\lim_{x \to -1^{-}} \frac{x}{x^{2}+x} = - \infty$

$\lim_{x \to -1^{+}} \frac{x}{x^{2}+x} = + \infty$

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:

$\frac{x}{x^{2}+ x} = \frac{x}{x\cdot (x + 1)} = \frac{1}{x + 1}$

$g(x) = \frac{1}{x + 1}$

The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$.

To learn more on lateral limits: brainly.com/question/21783151

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