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2 years ago

15

Today, a bicycle rider rode her bike 5 1/2 miles, she rode 6 1/4 miles. The difference in length between the two riders is what

fraction of the longer ride

1 answer:

wlad13 [49]2 years ago

7 0

Longer ridedifference and the answer

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3 years ago

Can you find the limits of this

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

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:

f(x) = cxⁿ
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

6 0

2 years ago