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# 1.5 Determining Limits Using Algebraic Properties of Limits

Anusha Tekumulla

### AP Calculus AB/BCΒ βΎοΈ

279Β resources
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π₯Watch: AP Calculus AB/BC - Algebraic Limits
This topic focuses on using the properties of limits to determine a limit value. We refer to these properties as algebraic because they are essentially the same properties you learn in algebra. Below, youβll find a list of properties of limits:

Medium.com

Here are some example problems to better understand the limit properties:Β

## Example Problem: Using the Sum Rule to Determine a Limit β

Explanation: In this example, we need to use the sum rule. Our first function is x^2 and our second function is x^3. Because the functions are being added together, we can evaluate their limits separately. The limit of x^2 as x approaches 3 is 9. The limit of x^3 as x approaches 3 is 27. Thus, the limit of (x^2 + x^3) as x approaches 3 is 9 + 27 = 36.Β The sum rule is very similar to the difference rule. If the function was x^2 - x^3, we would use the difference rule and the answer would be 9 - 27 = -18.Β

## Example Problem: Using the Constant Multiple Rule to Determine a Limit β‘

Explanation: In this example, we need to use the constant multiple rule. We can separate the constant (12) from the function (x^3). With the constant out of the way, we can solve the limit. The limit of x^3 as x approaches 5 is 125. Thus, the limit of (12x^3) as x approaches 5 is 125 * 12 = 1500.Β

## Example Problem: Using the Product Rule to Determine a Limit βοΈ

Explanation: In this example, we need to use the product rule. We can separate the first function (12x^3) from the second function (27x^-2). The limit of the first function as x approaches 5 is 1500 (from the previous example). The limit of the second function as x approaches 5 is 27 β 25 = 1.08. Thus the limit of (12x^3 * 27x^-2) as x approaches 5 is 1500 * 1.08 = 1620.Β
The difference rule is essentially the same as the product rule. If the function in this example were 12x^3 / 27x^-2, we would use the difference rule and the answer would be 1500 / 1.08 = 1388.889.Β

## Example Problem: Using the Exponent/Power Rule to Determine a Limit π₯

Explanation: In this example, we need to use the power rule. We can separate the function (12x3) from the exponent. The limit of the function as x approaches 5 is 9. Now, we just have to apply the exponent to the answer. Thus the limit of (x + 4)^3 as x approaches 5 is 9^3 = 729.Β
The root rule is essentially the same as the power rule. If the function in this example were β(x + 4), we would use the root rule and the answer would be β9 = 2.08
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πUnit 1 β Limits & Continuity
π€Unit 2 β Fundamentals of Differentiation
π€π½Unit 3 β Composite, Implicit, & Inverse Functions
πUnit 4 β Contextual Applications of Differentiation
β¨Unit 5 β Analytical Applications of Differentiation
π₯Unit 6 β Integration & Accumulation of Change
πUnit 7 β Differential Equations
πΆUnit 8 β Applications of Integration
π¦Unit 9 β Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)
βΎUnit 10 β Infinite Sequences & Series (BC Only)
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