- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Differentials
Problem 4.7.48
Textbook Question
17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.
lim_y→2 (y²+y-6) / (√(8-y²)-y)

1
First, substitute y = 2 into the expression to check if the limit results in an indeterminate form. You will find that both the numerator and the denominator evaluate to 0, indicating a 0/0 indeterminate form.
Since the limit is in an indeterminate form, l'Hôpital's Rule can be applied. According to l'Hôpital's Rule, if the limit of f(y)/g(y) as y approaches a value results in 0/0 or ∞/∞, then the limit can be evaluated as the limit of f'(y)/g'(y).
Differentiate the numerator, f(y) = y² + y - 6, to get f'(y) = 2y + 1.
Differentiate the denominator, g(y) = √(8-y²) - y. The derivative of √(8-y²) is (-y)/(√(8-y²)) using the chain rule, and the derivative of -y is -1. Therefore, g'(y) = (-y)/(√(8-y²)) - 1.
Apply l'Hôpital's Rule by taking the limit of the new fraction: lim_{y→2} (2y + 1) / ((-y)/(√(8-y²)) - 1). Substitute y = 2 into this expression to evaluate the limit.
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