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๐Ÿ‘‘ย  Unit 1: Limits & Continuity

๐Ÿถย  Unit 8: Applications of Integration

โ™พย  Unit 10: Infinite Sequences and Series (BC Only)

1.5 Determining Limits Using Algebraic Properties of Limits

#limits

#algebraicproperties

โฑ๏ธย ย 3 min read

written by

Anusha Tekumulla

anusha tekumulla


๐ŸŽฅWatch: 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, youโ€™ll find a list of properties of limits:

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

Example Problem: Using the Sum Rule to Determine a Limit โž•

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Answer: 36

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.ย 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.ย 

Example Problem: Using the Constant Multiple Rule to Determine a Limit โžก

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Answer: 1500

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.ย 

Example Problem: Using the Product Rule to Determine a Limit โœ–๏ธ

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Answer: 1620

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 โ„ 25 = 1.08. Thus the limit of (12x^3 * 27x^-2) as x approaches 5 is 1500 * 1.08 = 1620.ย 

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.ย 

Example Problem: Using the Exponent/Power Rule to Determine a Limit ๐Ÿ’ฅ

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Answer: 729

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.ย 

The root rule is essentially the same as the power rule. If the function in this example were โˆ›(x + 4), we would use the root rule and the answer would be โˆ›9 = 2.08

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