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Find
lim_(x rarr-2)(x^(3)+3x^(2)+2x)/(x+2).
Choose 1 answer:
(A) 6
(B) 0
(c) 2
(D) The limit doesn't exist

Find $\lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x+2}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $6$ $\newline$ (B) $0$ $\newline$ (C) $2$ $\newline$ (D) The limit doesn't exist

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

Full solution

Q. Find

\lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x+2}

.

\newline

Choose

1

answer:

\newline

(A)

6

\newline

(B)

0

\newline

(C)

2

\newline

(D) The limit doesn't exist

Check Substitution: We need to find the limit of the function as $x$ approaches $-2$ . Let's first try to directly substitute $x = -2$ into the function to see if the function is defined at that point. $\newline$ $\lim_{x \to -2}\frac{x^3 + 3x^2 + 2x}{x + 2} = \frac{(-2)^3 + 3(-2)^2 + 2(-2)}{-2 + 2}$ $\newline$ $= \frac{-8 + 12 - 4}{0}$ $\newline$ $= \frac{0}{0}$ $\newline$ This is an indeterminate form, which means we cannot directly find the limit by substitution.
Factorize and Simplify: Since we have an indeterminate form of $0/0$ , we should try to simplify the expression to see if we can cancel out the factor causing the indeterminate form. Let's factor the numerator. $\newline$ $x^3 + 3x^2 + 2x = x(x^2 + 3x + 2)$ $\newline$ Now we factor the quadratic $x^2 + 3x + 2$ . $\newline$ $x^2 + 3x + 2 = (x + 1)(x + 2)$ $\newline$ So the original function becomes: $\newline$ $(x(x + 1)(x + 2)) / (x + 2)$
Cancel Common Factor: We can now cancel out the $(x + 2)$ term in the numerator and the denominator, as long as $x$ is not equal to $-2$ (since division by zero is undefined). $\newline$ The simplified function is: $\newline$ $x(x + 1) = x^2 + x$ $\newline$ Now we can find the limit by direct substitution since the function is no longer undefined at $x = -2$ . $\newline$ $\lim_{x \to -2}(x^2 + x) = (-2)^2 + (-2) = 4 - 2 = 2$
Find Limit: The limit of the function as $x$ approaches $-2$ is $2$ .

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