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rjkz [21]
2 years ago
15

PLEASE HELP WILL MARK BRAINLEIST

Mathematics
1 answer:
Bumek [7]2 years ago
8 0

Answer:

I belive it's Y = 3/2x - 1

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The answer you are looking for is the letter B on your assignment hun. Merry almost Christmas☃️

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Can you find the limits of this ​
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Answer:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Constant]:                                                                                             \displaystyle \lim_{x \to c} b = b

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Property [Addition/Subtraction]:                                                                   \displaystyle \lim_{x \to c} [f(x) \pm g(x)] =  \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}

Let's substitute in <em>x</em> = -2 using the limit rule:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}

Evaluating this, we arrive at an indeterminate form:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}

Substitute in <em>x</em> = -2 using the limit rule:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}

Evaluating this, we get:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Limits

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By the way, this is copied from the internet.

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