Answer:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x} = \frac{-1}{6}[/tex]
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Constant]: [tex]\displaystyle \lim_{x \to c} b = b[/tex]
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Limit Property [Addition/Subtraction]: [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]
L'Hopital's Rule: [tex]\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}[/tex]
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
We are given the following limit:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x}[/tex]
Substitute in x = 9 using the limit rule:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x} = \frac{\sqrt{9} - 3}{9 - 9}[/tex]
Evaluating this, we have an indeterminate form:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x} = \frac{0}{0}[/tex]
Since we have an indeterminate form, let's use L'Hopital's Rule:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x} = \lim_{x \to 9} \frac{\frac{1}{2\sqrt{x}}}{-1}[/tex]
Simplify:
[tex]\displaystyle \lim_{x \to 9} \frac{\sqrt{x} - 3}{9 - x} = \lim_{x \to 9} \frac{-1}{2\sqrt{x}}[/tex]
Substitute in x = 9 using the limit rule:
[tex]\displaystyle \lim_{x \to 9} \frac{-1}{2\sqrt{x}} = \frac{-1}{2\sqrt{9}}[/tex]
Evaluating this, we get:
[tex]\displaystyle \lim_{x \to 9} \frac{-1}{2\sqrt{x}} = \frac{-1}{6}[/tex]
And we have our answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
