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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. Substitute and Evaluate: First, let's try to directly substitute the value of xx with 4-4 into the function to see if it results in an indeterminate form or not.\newlineSubstitute x=4x = -4 into (7x+28)/(x2+x12)(7x+28)/(x^2+x-12).\newlinelimx4(7x+28)/(x2+x12)\lim_{x \rightarrow -4}(7x+28)/(x^2+x-12)\newline= (7(4)+28)/((4)2+(4)12)(7(-4)+28)/((-4)^2+(-4)-12)\newline= (28+28)/(16412)(-28+28)/(16-4-12)\newline= 0/00/0\newlineSince we get 0/00/0, which is an indeterminate form, we cannot directly evaluate the limit by substitution.
  2. Factorize Denominator: Next, we should try to factor the numerator and the denominator to see if there are any common factors that can be canceled out.\newlineFactor the quadratic expression in the denominator x2+x12x^2 + x - 12.\newlinex2+x12x^2 + x - 12 can be factored into (x+4)(x3)(x + 4)(x - 3).\newlineNow, let's rewrite the limit expression with the factored form of the denominator.\newlinelimx47x+28(x+4)(x3)\lim_{x \rightarrow -4}\frac{7x+28}{(x + 4)(x - 3)}
  3. Factorize Numerator: Notice that the numerator 7x+287x + 28 can also be factored because it is a common factor of 77.\newlineFactor out the common factor of 77 from the numerator.\newline7x+287x + 28 can be factored into 7(x+4)7(x + 4).\newlineNow, let's rewrite the limit expression with the factored form of the numerator.\newlinelimx47(x+4)(x+4)(x3)\lim_{x \rightarrow -4}\frac{7(x + 4)}{(x + 4)(x - 3)}
  4. Cancel Common Factor: We can now cancel out the common factor (x+4)(x + 4) from the numerator and the denominator, as long as xx is not equal to 4-4. Since we are taking the limit as xx approaches 4-4, not at x=4x = -4, we can cancel the factors.\newlineCancel the (x+4)(x + 4) term from the numerator and the denominator.\newlinelimx47(x3)\lim_{x \to -4}\frac{7}{(x - 3)}
  5. Final Substitution: Now that we have a simplified expression, we can substitute x=4x = -4 into the limit to find the value.\newlineSubstitute x=4x = -4 into 7(x3)\frac{7}{(x - 3)}.\newlinelimx47(x3)\lim_{x \rightarrow -4}\frac{7}{(x - 3)}\newline= 7((4)3)\frac{7}{((-4) - 3)}\newline= 7(7)\frac{7}{(-7)}\newline= 1-1\newlineThe limit exists and is equal to 1-1.

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