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

Check for Indeterminate Form: Identify if direct substitution of x = -2 into the function gives an indeterminate form. Let's substitute x = -2 into the function and see what we get. lim_(x -> -2)(x^3 + 3x^2 + 2x) / (x + 2) = ((-2)^3 + 3(-2)^2 + 2(-2)) / (-2 + 2) = (-8 + 12 - 4) / 0 = 0 / 0 This is an indeterminate form, so we cannot simply substitute x = -2 to find the limit.Factorize the Numerator: Factor the numerator to simplify the expression. We notice that the numerator is a polynomial that can be factored. Let's try to factor it. x^3 + 3x^2 + 2x = x(x^2 + 3x + 2) Now, let's factor the quadratic part of the expression. x^2 + 3x + 2 can be factored into (x + 1)(x + 2). So the entire expression becomes: x(x + 1)(x + 2)Cancel Common Factor: Cancel out the common factor from the numerator and the denominator. We have a common factor of (x + 2) in both the numerator and the denominator. Let's cancel it out. lim_(x -> -2)(x(x + 1)(x + 2)) / (x + 2) After canceling out the common factor, we get: lim_(x -> -2)x(x + 1)Final Direct Substitution: Now that we have simplified the expression, let's try direct substitution again. lim_(x -> -2)x(x + 1) = (-2)(-2 + 1) = (-2)(-1) = 2

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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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Calculus

Find derivatives of logarithmic functions

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 for Indeterminate Form: Identify if direct substitution of $x = -2$ into the function gives an indeterminate form. $\newline$ Let's substitute $x = -2$ into the function and see what we get. $\newline$ $\lim_{x \to -2}\frac{x^3 + 3x^2 + 2x}{x + 2} = \frac{(-2)^3 + 3(-2)^2 + 2(-2)}{-2 + 2}$ $\newline$ $= \frac{-8 + 12 - 4}{0}$ $\newline$ $= \frac{0}{0}$ $\newline$ This is an indeterminate form, so we cannot simply substitute $x = -2$ to find the limit.
Factorize the Numerator: Factor the numerator to simplify the expression. $\newline$ We notice that the numerator is a polynomial that can be factored. Let's try to factor it. $\newline$ $x^3 + 3x^2 + 2x = x(x^2 + 3x + 2)$ $\newline$ Now, let's factor the quadratic part of the expression. $\newline$ $x^2 + 3x + 2$ can be factored into $(x + 1)(x + 2)$ . $\newline$ So the entire expression becomes: $\newline$ $x(x + 1)(x + 2)$
Cancel Common Factor: Cancel out the common factor from the numerator and the denominator. $\newline$ We have a common factor of $(x + 2)$ in both the numerator and the denominator. Let's cancel it out. $\newline$ $\lim_{x \to -2}\frac{x(x + 1)(x + 2)}{x + 2}$ $\newline$ After canceling out the common factor, we get: $\newline$ $\lim_{x \to -2}x(x + 1)$
Final Direct Substitution: Now that we have simplified the expression, let's try direct substitution again. $\lim_{x \to -2}x(x + 1) = (-2)(-2 + 1)$ = $(-2)(-1)$ = $2$