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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 2, Problem 11

Determine the following limits.
lim x→1 √5x+6

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1
Identify the type of limit: This is a direct substitution limit problem where you need to evaluate the limit of a function as x approaches a specific value.
Substitute the value x = 1 directly into the function: \( \lim_{x \to 1} \sqrt{5x + 6} \).
Calculate the expression inside the square root: Substitute x = 1 into the expression 5x + 6 to get 5(1) + 6.
Simplify the expression: Calculate 5(1) + 6 to get the value inside the square root.
Evaluate the square root: Take the square root of the simplified expression to find the limit.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding how functions behave near specific points, which is crucial for defining derivatives and integrals. In this case, we are interested in the limit of the function as x approaches 1.
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Continuous Functions

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. Continuous functions do not have breaks, jumps, or holes, making it easier to evaluate limits. The function √(5x + 6) is continuous at x = 1, allowing us to directly substitute the value into the function.
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Substitution in Limits

Substitution is a technique used in evaluating limits where you replace the variable in the function with the value that the variable is approaching. If the function is continuous at that point, this method yields the limit directly. For the limit lim x→1 √(5x + 6), we can substitute x = 1 into the function to find the limit.
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