Find lim_(x rarr-2)(3-sqrt(6x+21))/(x+2). Choose 1 answer: (A) -1 (B) -2 (c) -3 (D) The limit doesn't exist

Substitute and Indeterminate Form: First, let's try to directly substitute the value of x = -2 into the limit expression to see if it results in an indeterminate form. lim_(x -> -2)(3 - sqrt(6x + 21)) / (x + 2) = (3 - sqrt(6(-2) + 21)) / (-2 + 2) = (3 - sqrt(-12 + 21)) / 0 = (3 - sqrt(9)) / 0 = (3 - 3) / 0 = 0 / 0 This is an indeterminate form, so we cannot directly evaluate the limit by substitution.Algebraic Manipulation: Since we have an indeterminate form of 0/0, we need to use algebraic manipulation to simplify the expression and eliminate the indeterminate form. We can multiply the numerator and the denominator by the conjugate of the numerator to rationalize it. The conjugate of the numerator 3 - sqrt(6x + 21) is 3 + sqrt(6x + 21). Let's multiply the numerator and the denominator by this conjugate. lim_(x -> -2)((3 - sqrt(6x + 21))(3 + sqrt(6x + 21))) / ((x + 2)(3 + sqrt(6x + 21)))Numerator Multiplication: Now, let's perform the multiplication in the numerator, which is a difference of squares. (3 - sqrt(6x + 21))(3 + sqrt(6x + 21)) = 3^2 - (sqrt(6x + 21))^2 = 9 - (6x + 21) = 9 - 6x - 21 = -6x - 12Factor Cancelation: We can now simplify the expression by canceling out the common factor of (x + 2) in the numerator and the denominator. lim_(x -> -2)(-6x - 12) / ((x + 2)(3 + sqrt(6x + 21))) = lim_(x -> -2)(-6(x + 2)) / ((x + 2)(3 + sqrt(6x + 21))) = lim_(x -> -2)(-6) / (3 + sqrt(6x + 21))Final Limit Calculation: Now that the indeterminate form is eliminated, we can substitute x = -2 into the simplified expression to find the limit. lim_(x -> -2)(-6) / (3 + sqrt(6x + 21)) = (-6) / (3 + sqrt(6(-2) + 21)) = (-6) / (3 + sqrt(-12 + 21)) = (-6) / (3 + sqrt(9)) = (-6) / (3 + 3) = (-6) / 6 = -1

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Find
lim_(x rarr-2)(3-sqrt(6x+21))/(x+2).
Choose 1 answer:
(A) -1
(B) -2
(c) -3
(D) The limit doesn't exist

Find $\lim _{x \rightarrow-2} \frac{3-\sqrt{6 x+21}}{x+2}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-1$ $\newline$ (B) $-2$ $\newline$ (C) $-3$ $\newline$ (D) The limit doesn't exist

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Algebra 1

Is (x, y) a solution to the system of equations?

Full solution

Q. Find

\lim _{x \rightarrow-2} \frac{3-\sqrt{6 x+21}}{x+2}

\newline

Choose

1

answer:

\newline

(A)

-1

\newline

(B)

-2

\newline

(C)

-3

\newline

(D) The limit doesn't exist

Substitute and Indeterminate Form: First, let's try to directly substitute the value of $x = -2$ into the limit expression to see if it results in an indeterminate form. $\lim_{x \to -2}\frac{3 - \sqrt{6x + 21}}{x + 2}$ = $\frac{3 - \sqrt{6(-2) + 21}}{-2 + 2}$ = $\frac{3 - \sqrt{-12 + 21}}{0}$ = $\frac{3 - \sqrt{9}}{0}$ = $\frac{3 - 3}{0}$ = $\frac{0}{0}$ This is an indeterminate form, so we cannot directly evaluate the limit by substitution.
Algebraic Manipulation: Since we have an indeterminate form of $0/0$ , we need to use algebraic manipulation to simplify the expression and eliminate the indeterminate form. We can multiply the numerator and the denominator by the conjugate of the numerator to rationalize it. $\newline$ The conjugate of the numerator $3 - \sqrt{6x + 21}$ is $3 + \sqrt{6x + 21}$ . Let's multiply the numerator and the denominator by this conjugate. $\newline$ $\lim_{x \to -2}\frac{(3 - \sqrt{6x + 21})(3 + \sqrt{6x + 21})}{(x + 2)(3 + \sqrt{6x + 21})}$
Numerator Multiplication: Now, let's perform the multiplication in the numerator, which is a difference of squares. $\newline$ $(3 - \sqrt{6x + 21})(3 + \sqrt{6x + 21}) = 3^2 - (\sqrt{6x + 21})^2$ $\newline$ $= 9 - (6x + 21)$ $\newline$ $= 9 - 6x - 21$ $\newline$ $= -6x - 12$
Factor Cancelation: We can now simplify the expression by canceling out the common factor of $(x + 2)$ in the numerator and the denominator. $\newline$ $\lim_{x \to -2}\frac{-6x - 12}{(x + 2)(3 + \sqrt{6x + 21})}$ $\newline$ = $\lim_{x \to -2}\frac{-6(x + 2)}{(x + 2)(3 + \sqrt{6x + 21})}$ $\newline$ = $\lim_{x \to -2}\frac{-6}{3 + \sqrt{6x + 21}}$
Final Limit Calculation: Now that the indeterminate form is eliminated, we can substitute $x = -2$ into the simplified expression to find the limit. $\lim_{x \to -2}\frac{-6}{3 + \sqrt{6x + 21}}$ = $\frac{-6}{3 + \sqrt{6(-2) + 21}}$ = $\frac{-6}{3 + \sqrt{-12 + 21}}$ = $\frac{-6}{3 + \sqrt{9}}$ = $\frac{-6}{3 + 3}$ = $\frac{-6}{6}$ = $-1$