Find lim_(x rarr-4)(x^(2)-15)/(x^(2)-16). Choose 1 answer: (A) -(1)/(8) (B) (1)/(8) (C) (19)/(20) (D) The limit doesn't exist

Substitute x = -4: We are asked to find the limit of the function (x^2 - 15) / (x^2 - 16) as x approaches -4. Let's first try to directly substitute x = -4 into the function to see if the limit can be computed this way.Calculate numerator and denominator: Substitute x = -4 into the function: lim_(x -> -4)(x^2 - 15) / (x^2 - 16) = ((-4)^2 - 15) / ((-4)^2 - 16)Factor the denominator: Calculate the numerator and the denominator separately: Numerator: (-4)^2 - 15 = 16 - 15 = 1 Denominator: (-4)^2 - 16 = 16 - 16 = 0Rewrite the limit expression: We notice that the denominator becomes 0, which means the function is undefined at x = -4. However, this does not necessarily mean that the limit does not exist. We need to factor the denominator to see if there is a common factor that can be canceled out with the numerator.Conclusion: Factor the denominator: x^2 - 16 can be factored as (x - 4)(x + 4).Now, let's rewrite the limit expression with the factored denominator: lim_(x -> -4)(x^2 - 15) / ((x - 4)(x + 4))We see that there is no common factor between the numerator and the denominator that can be canceled out. Since the denominator is zero and the numerator is non-zero when x = -4, the limit does not exist.

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Find
lim_(x rarr-4)(x^(2)-15)/(x^(2)-16).
Choose 1 answer:
(A)
-(1)/(8)
(B)
(1)/(8)
(C)
(19)/(20)
(D) The limit doesn't exist

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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Full solution

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

Substitute $x = -4$ : We are asked to find the limit of the function $\frac{x^2 - 15}{x^2 - 16}$ as $x$ approaches $-4$ . Let's first try to directly substitute $x = -4$ into the function to see if the limit can be computed this way.
Calculate numerator and denominator: Substitute $x = -4$ into the function: $\lim_{x \to -4}\frac{x^2 - 15}{x^2 - 16} = \frac{(-4)^2 - 15}{(-4)^2 - 16}$
Factor the denominator: Calculate the numerator and the denominator separately: $\newline$ Numerator: $(-4)^2 - 15 = 16 - 15 = 1$ $\newline$ Denominator: $(-4)^2 - 16 = 16 - 16 = 0$
Rewrite the limit expression: We notice that the denominator becomes $0$ , which means the function is undefined at $x = -4$ . However, this does not necessarily mean that the limit does not exist. We need to factor the denominator to see if there is a common factor that can be canceled out with the numerator.
Conclusion: Factor the denominator: $x^2 - 16$ can be factored as $(x - 4)(x + 4)$ .
Conclusion: Factor the denominator: $\newline$ $x^2 - 16$ can be factored as $(x - 4)(x + 4)$ .Now, let's rewrite the limit expression with the factored denominator: $\newline$ $\lim_{x \to -4}\frac{x^2 - 15}{(x - 4)(x + 4)}$
Conclusion: Factor the denominator: $\newline$ $x^2 - 16$ can be factored as $(x - 4)(x + 4)$ .Now, let's rewrite the limit expression with the factored denominator: $\newline$ $\lim_{x \to -4}\frac{x^2 - 15}{(x - 4)(x + 4)}$ We see that there is no common factor between the numerator and the denominator that can be canceled out. Since the denominator is zero and the numerator is non-zero when $x = -4$ , the limit does not exist.