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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 limx2x3+3x2+2xx+2 \lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x+2} .\newlineChoose 11 answer:\newline(A) 66\newline(B) 00\newline(C) 22\newline(D) The limit doesn't exist

Full solution

Q. Find limx2x3+3x2+2xx+2 \lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x+2} .\newlineChoose 11 answer:\newline(A) 66\newline(B) 00\newline(C) 22\newline(D) The limit doesn't exist
  1. Check for Indeterminate Form: Identify if direct substitution of x=2x = -2 into the function gives an indeterminate form.\newlineLet's substitute x=2x = -2 into the function and see what we get.\newlinelimx2x3+3x2+2xx+2=(2)3+3(2)2+2(2)2+2\lim_{x \to -2}\frac{x^3 + 3x^2 + 2x}{x + 2} = \frac{(-2)^3 + 3(-2)^2 + 2(-2)}{-2 + 2}\newline=8+1240= \frac{-8 + 12 - 4}{0}\newline=00= \frac{0}{0}\newlineThis is an indeterminate form, so we cannot simply substitute x=2x = -2 to find the limit.
  2. Factorize the Numerator: Factor the numerator to simplify the expression.\newlineWe notice that the numerator is a polynomial that can be factored. Let's try to factor it.\newlinex3+3x2+2x=x(x2+3x+2)x^3 + 3x^2 + 2x = x(x^2 + 3x + 2)\newlineNow, let's factor the quadratic part of the expression.\newlinex2+3x+2x^2 + 3x + 2 can be factored into (x+1)(x+2)(x + 1)(x + 2).\newlineSo the entire expression becomes:\newlinex(x+1)(x+2)x(x + 1)(x + 2)
  3. Cancel Common Factor: Cancel out the common factor from the numerator and the denominator.\newlineWe have a common factor of (x+2)(x + 2) in both the numerator and the denominator. Let's cancel it out.\newlinelimx2x(x+1)(x+2)x+2\lim_{x \to -2}\frac{x(x + 1)(x + 2)}{x + 2}\newlineAfter canceling out the common factor, we get:\newlinelimx2x(x+1)\lim_{x \to -2}x(x + 1)
  4. Final Direct Substitution: Now that we have simplified the expression, let's try direct substitution again. limx2x(x+1)=(2)(2+1)\lim_{x \to -2}x(x + 1) = (-2)(-2 + 1) = (2)(1)(-2)(-1) = 22

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