Find $\lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16}$ $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{8}$ $\newline$ (B) $\frac{1}{8}$ $\newline$ (c) $\frac{19}{20}$ $\newline$ (D) The limit doesn't exist

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Algebra 1

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Q. Find

\lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16}

\newline

Choose

1

answer:

\newline

(A)

-\frac{1}{8}

\newline

(B)

\frac{1}{8}

\newline

(c)

\frac{19}{20}

\newline

(D) The limit doesn't exist

Identify the limit form: Identify the form of the limit. $\newline$ We need to find the limit of the function as $x$ approaches $-4$ . Let's substitute $x = -4$ into the function to see if we can directly evaluate the limit. $\newline$ $f(x) = \frac{x^2 - 15}{x^2 - 16}$ $\newline$ $f(-4) = \frac{(-4)^2 - 15}{(-4)^2 - 16}$ $\newline$ $f(-4) = \frac{16 - 15}{16 - 16}$ $\newline$ $f(-4) = \frac{1}{0}$ $\newline$ This results in a division by zero, which is undefined. Therefore, we cannot directly evaluate the limit by substitution.
Substitute $x = -4$ : Factor the denominator and simplify if possible. $\newline$ The denominator $x^2 - 16$ is a difference of squares and can be factored as $(x + 4)(x - 4)$ . $\newline$ $f(x) = \frac{x^2 - 15}{(x + 4)(x - 4)}$ $\newline$ Since we are looking for the limit as $x$ approaches $-4$ , we can check if the numerator has a factor of $(x + 4)$ that would cancel out the denominator's $(x + 4)$ . $\newline$ However, the numerator $x^2 - 15$ does not factor with $x + 4$ as a factor, so we cannot simplify the function by canceling out terms.
Division by zero: Evaluate the limit using algebraic manipulation. $\newline$ Since we cannot simplify the function by canceling out terms, we need to use another method to evaluate the limit. We can use the fact that the function is continuous everywhere except at $x = 4$ and $x = -4$ (where the denominator is zero). $\newline$ For values of $x$ close to but not equal to $-4$ , the function is defined and continuous. Therefore, we can evaluate the limit by approaching $-4$ from either side. $\newline$ As $x$ approaches $-4$ , the numerator approaches $(-4)^2 - 15 = 16 - 15 = 1$ , and the denominator approaches $(-4)^2 - 16 = 16 - 16 = 0$ , but since we are taking a limit, we are not actually dividing by zero. $\newline$ The limit of the numerator as $x$ approaches $-4$ is $x = -4$ $1$ , and the limit of the denominator as $x$ approaches $-4$ is a very small number close to zero but not equal to zero. Therefore, the limit of the function as $x$ approaches $-4$ is a very large number. $\newline$ Since the denominator is $x = -4$ $6$ , and we are approaching $-4$ , the sign of the denominator is positive (because $x = -4$ $8$ ). Therefore, the limit is positive. $\newline$ The limit of the function as $x$ approaches $-4$ is positive infinity.