Question

Need help finding the limits

Answer 1

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x) = (-1)/(6)`

General Formulas and Concepts:

Calculus

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:
`\displaystyle \lim_(x \to c) (f(x))/(g(x)) = \lim_(x \to c) (f'(x))/(g'(x))`

Differentiation

• Derivatives
• Derivative Notation

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

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Explanation:

We are given the following limit:

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x)`

Substitute in x = 9 using the limit rule:

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x) = (√(9) - 3)/(9 - 9)`

Evaluating this, we have an indeterminate form:

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x) = (0)/(0)`

Since we have an indeterminate form, let's use L'Hopital's Rule:

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x) = \lim_(x \to 9) ((1)/(2√(x)))/(-1)`

Simplify:

`\displaystyle \lim_(x \to 9) (√(x) - 3)/(9 - x) = \lim_(x \to 9) (-1)/(2√(x))`

Substitute in x = 9 using the limit rule:

`\displaystyle \lim_(x \to 9) (-1)/(2√(x)) = (-1)/(2√(9))`

Evaluating this, we get:

`\displaystyle \lim_(x \to 9) (-1)/(2√(x)) = (-1)/(6)`

And we have our answer.

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

Unit: Limits