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

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Find derivatives of logarithmic functions

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Q. Find

\lim _{x \rightarrow-4} \frac{7 x+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

Substitution Attempt: First, let's try to directly substitute the value of $x$ into the expression to see if the limit can be evaluated this way. $\newline$ $\lim_{x \to -4}\frac{7x+28}{x^2+x-12}$ $\newline$ $= \frac{7(-4)+28}{(-4)^2-4-12}$ $\newline$ $= \frac{(-28+28)}{16-4-12}$ $\newline$ $= \frac{0}{0}$ $\newline$ We get an indeterminate form $0/0$ , which means we need to use a different method to find the limit.
Factoring the Denominator: Since we have an indeterminate form, we should try to factor the quadratic expression in the denominator and see if there is a common factor that can be canceled with the numerator. $\newline$ The quadratic expression $x^2 + x - 12$ can be factored into $(x+4)(x-3)$ .
Rewriting with Factored Denominator: Now let's rewrite the limit expression with the factored denominator and see if we can simplify it. $\newline$ $\lim_{x \to -4}\frac{7x+28}{(x+4)(x-3)}$ $\newline$ = $\lim_{x \to -4}\frac{7(x+4)}{(x+4)(x-3)}$ $\newline$ Notice that $(x+4)$ is a common factor in both the numerator and the denominator.
Canceling Common Factor: We can now cancel the common factor $(x+4)$ from the numerator and the denominator. $\newline$ $\lim_{x \to -4}\frac{7(x+4)}{(x+4)(x-3)}$ $\newline$ $= \lim_{x \to -4}\frac{7}{x-3}$ $\newline$ Now that the expression is simplified, we can directly substitute $x = -4$ into the limit.
Direct Substitution: Substitute $x = -4$ into the simplified expression to find the limit. $\lim_{x \to -4}\frac{7}{(x-3)}$ = \frac{ $7$ }{(( $-4$ ) $-3$ )}= \frac{ $7$ }{( $-7$ )}= $-1$ We have found the limit, and it exists.