Find lim_(x rarr2)(x^(4)-4x^(3)+4x^(2))/(x-2). Choose 1 answer: (A) -4 (B) 0 (C) 4 (D) The limit doesn't exist

Identify the form: Identify the form of the limit. We need to find the limit of the function as x approaches 2. Let's substitute x = 2 into the function to see what form the limit takes. lim_(x -> 2)(x^4 - 4x^3 + 4x^2) / (x - 2) = (2^4 - 4*2^3 + 4*2^2) / (2 - 2) = (16 - 32 + 16) / 0 = 0 / 0 This is an indeterminate form, which means we need to do further work to find the limit.Factor the numerator: Factor the numerator. Since the direct substitution gives us an indeterminate form, we can try to factor the numerator to see if we can simplify the expression. The numerator is a polynomial that seems to be a perfect square trinomial. Let's factor it: x^4 - 4x^3 + 4x^2 = (x^2)^2 - 2*2*x^2*x + (2x)^2 = (x^2 - 2x)^2 Now we have: lim_(x -> 2)(x^2 - 2x)^2 / (x - 2)Factor out a term: Factor out a term of (x - 2) from the numerator. We notice that (x^2 - 2x) can be further factored to x(x - 2). This will allow us to cancel out the (x - 2) term in the denominator. (x^2 - 2x)^2 = (x(x - 2))^2 = x^2 * (x - 2)^2 Now we have: lim_(x -> 2)x^2 * (x - 2)^2 / (x - 2)Cancel the common term: Cancel the common term. We can now cancel the common (x - 2) term from the numerator and the denominator. lim_(x -> 2)x^2 * (x - 2)^2 / (x - 2) = lim_(x -> 2)x^2 * (x - 2)Evaluate the limit: Evaluate the limit. Now that we have simplified the expression, we can substitute x = 2 to find the limit. lim_(x -> 2)x^2 * (x - 2) = 2^2 * (2 - 2) = 4 * 0 = 0

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Find
lim_(x rarr2)(x^(4)-4x^(3)+4x^(2))/(x-2).
Choose 1 answer:
(A) -4
(B) 0
(C) 4
(D) The limit doesn't exist

Find $\lim _{x \rightarrow 2} \frac{x^{4}-4 x^{3}+4 x^{2}}{x-2}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-4$ $\newline$ (B) $0$ $\newline$ (C) $4$ $\newline$ (D) The limit doesn't exist

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Algebra 2

Solve quadratic inequalities

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Q. Find

\lim _{x \rightarrow 2} \frac{x^{4}-4 x^{3}+4 x^{2}}{x-2}

\newline

Choose

1

answer:

\newline

(A)

-4

\newline

(B)

0

\newline

(C)

4

\newline

(D) The limit doesn't exist

Identify the form: Identify the form of the limit. $\newline$ We need to find the limit of the function as $x$ approaches $2$ . Let's substitute $x = 2$ into the function to see what form the limit takes. $\newline$ $\lim_{x \to 2}\frac{x^4 - 4x^3 + 4x^2}{x - 2} = \frac{2^4 - 4\cdot2^3 + 4\cdot2^2}{2 - 2}$ $\newline$ $= \frac{16 - 32 + 16}{0}$ $\newline$ $= \frac{0}{0}$ $\newline$ This is an indeterminate form, which means we need to do further work to find the limit.
Factor the numerator: Factor the numerator. $\newline$ Since the direct substitution gives us an indeterminate form, we can try to factor the numerator to see if we can simplify the expression. $\newline$ The numerator is a polynomial that seems to be a perfect square trinomial. Let's factor it: $\newline$ $x^4 - 4x^3 + 4x^2 = (x^2)^2 - 2\cdot 2\cdot x^2\cdot x + (2x)^2$ $\newline$ $= (x^2 - 2x)^2$ $\newline$ Now we have: $\newline$ $\lim_{x \to 2}\frac{(x^2 - 2x)^2}{(x - 2)}$
Factor out a term: Factor out a term of $(x - 2)$ from the numerator. $\newline$ We notice that $(x^2 - 2x)$ can be further factored to $x(x - 2)$ . This will allow us to cancel out the $(x - 2)$ term in the denominator. $\newline$ $(x^2 - 2x)^2 = (x(x - 2))^2$ $\newline$ $= x^2 * (x - 2)^2$ $\newline$ Now we have: $\newline$ $\lim_{x \to 2}x^2 * (x - 2)^2 / (x - 2)$
Cancel the common term: Cancel the common term. $\newline$ We can now cancel the common $x - 2$ term from the numerator and the denominator. $\newline$ $\lim_{x \to 2}x^2 * (x - 2)^2 / (x - 2) = \lim_{x \to 2}x^2 * (x - 2)$
Evaluate the limit: Evaluate the limit. $\newline$ Now that we have simplified the expression, we can substitute $x = 2$ to find the limit. $\newline$ $\lim_{x \to 2}x^2 * (x - 2) = 2^2 * (2 - 2)$ $\newline$ $= 4 * 0$ $\newline$ $= 0$