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Find
lim_(x rarr-4)(7x+28)/(x^(2)+x-12)
Choose 1 answer:
(A) 1
(B) 7
(C) -1
(D) The limit doesn't exist

Find $\lim_{x \to -4}\frac{7x+28}{x^{2}+x-12}$ $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $7$ $\newline$ (C) $-1$ $\newline$ (D) The limit doesn't exist

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Find limits of polynomials and rational functions

Full solution

Q. Find

\lim_{x \to -4}\frac{7x+28}{x^{2}+x-12}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

7

\newline

(C)

-1

\newline

(D) The limit doesn't exist

Substitute x Value: First, let's try to directly substitute the value of $x$ into the expression to see if the limit can be evaluated this way. $\newline$ $\lim_{x \to -4}\frac{7x+28}{x^2+x-12}$ $\newline$ Substitute $x = -4$ : $\newline$ $\frac{7(-4)+28}{(-4)^2+(-4)-12}$
Perform Calculations: Now, let's perform the calculations: $\newline$ $(7(-4)+28)/((16-4)-12)$ $\newline$ $(-28+28)/(16-4-12)$ $\newline$ $0/0$ $\newline$ We get an indeterminate form $0/0$ , which means we need to simplify the expression further to find the limit.
Factor Quadratic Polynomial: To simplify the expression, we can factor the quadratic polynomial in the denominator. $\newline$ The quadratic $x^2 + x - 12$ can be factored into $(x+4)(x-3)$ . $\newline$ So, we rewrite the limit expression as: $\newline$ $\lim_{x \to -4}\frac{7x+28}{(x+4)(x-3)}$
Factor Numerator: Notice that the numerator $7x+28$ can also be factored because it is a multiple of $7$ : $\newline$ $7x+28 = 7(x+4)$ $\newline$ Now the limit expression becomes: $\newline$ $\lim_{x \to -4}\frac{7(x+4)}{(x+4)(x-3)}$
Cancel Common Factor: We can now cancel out the common factor $(x+4)$ from the numerator and the denominator: $\lim_{x \to -4}\frac{7}{(x-3)}$
Substitute x Value: With the common factor canceled, we can now substitute $x = -4$ directly into the simplified expression: $\newline$ $\frac{7}{((-4)-3)}$ $\newline$ $\frac{7}{(-7)}$ $\newline$ $-1$
Final Answer: The limit of the expression as $x$ approaches $-4$ is $-1$ . Therefore, the correct answer is: (C) $-1$

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