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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 limx47x+28x2+x12 \lim _{x \rightarrow-4} \frac{7 x+28}{x^{2}+x-12} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 77\newline(C) 1-1\newline(D) The limit doesn't exist

Full solution

Q. Find limx47x+28x2+x12 \lim _{x \rightarrow-4} \frac{7 x+28}{x^{2}+x-12} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 77\newline(C) 1-1\newline(D) The limit doesn't exist
  1. Substitution Attempt: First, let's try to directly substitute the value of xx into the expression to see if the limit can be evaluated this way.\newlinelimx47x+28x2+x12\lim_{x \to -4}\frac{7x+28}{x^2+x-12}\newline=7(4)+28(4)2412= \frac{7(-4)+28}{(-4)^2-4-12}\newline=(28+28)16412= \frac{(-28+28)}{16-4-12}\newline=00= \frac{0}{0}\newlineWe get an indeterminate form 0/00/0, which means we need to use a different method to find the limit.
  2. 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.\newlineThe quadratic expression x2+x12x^2 + x - 12 can be factored into (x+4)(x3)(x+4)(x-3).
  3. Rewriting with Factored Denominator: Now let's rewrite the limit expression with the factored denominator and see if we can simplify it. \newlinelimx47x+28(x+4)(x3)\lim_{x \to -4}\frac{7x+28}{(x+4)(x-3)} \newline= limx47(x+4)(x+4)(x3)\lim_{x \to -4}\frac{7(x+4)}{(x+4)(x-3)} \newlineNotice that (x+4)(x+4) is a common factor in both the numerator and the denominator.
  4. Canceling Common Factor: We can now cancel the common factor (x+4)(x+4) from the numerator and the denominator.\newlinelimx47(x+4)(x+4)(x3)\lim_{x \to -4}\frac{7(x+4)}{(x+4)(x-3)}\newline=limx47x3= \lim_{x \to -4}\frac{7}{x-3}\newlineNow that the expression is simplified, we can directly substitute x=4x = -4 into the limit.
  5. Direct Substitution: Substitute x=4x = -4 into the simplified expression to find the limit.limx47(x3)\lim_{x \to -4}\frac{7}{(x-3)}= \frac{77}{((4-4)3-3)}= \frac{77}{(7-7)}= 1-1We have found the limit, and it exists.

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