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Continuity at a point Determine whether the following functions are continuous at a. Use the continuity checklist to justify your answer. $$f(x)=\left\\{\begin{array}{ll}\frac{x^{2}-1}{x-1} & \text { if } x \neq 1 \\\ 3 & \text { if } x=1\end{array} ; a=1\right.$$

Short Answer

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#Step_title# Answer #Answer_content# The given function is not continuous at x=1.

Step by step solution

01

Check if f(a) is defined for x=1

Since there is a specific definition for f(x) when x=1, we can say that f(a) is defined for x=1. In this case, $$f(1)=3$$
02

Find the limit as x approaches 1 of f(x)

As x approaches 1, we need to evaluate the expression: $$\lim_{x \to 1}\frac{x^2-1}{x-1}$$ This limit can be simplified before we evaluate it. Notice that the numerator is a difference of squares: $$\lim_{x \to 1}\frac{(x-1)(x+1)}{x-1}$$ We can cancel out the (x-1) from the numerator and the denominator: $$\lim_{x \to 1}(x+1)$$ Now, we can substitute x=1 to find the limit: $$\lim_{x \to 1}(1+1)=2$$
03

Check if the limit equals f(a)

In this case, we have the following information: 1. f(1) = 3 2. The limit as x approaches 1 of f(x) is 2 Since the limit as x approaches 1 of f(x) is not equal to f(1), the function is not continuous at x=1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits of Functions
Understanding the limit of a function is essential for analyzing function behavior as it approaches a point on the graph. This concept is particularly pivotal when ascertaining continuity of a function at a point.

For example, consider the expression \( \lim\text{{_{x \to a}}} f(x) \), which denotes the value that the function \( f(x) \) approaches as \( x \) approaches \( a \). In mathematical terms, if as \( x \) gets arbitrarily close to \( a \) from both sides, the function's value approaches some number \( L \), then \( L \) is said to be the limit of \( f(x) \) as \( x \) approaches \( a \).

It's useful to recognize that limits can often be found by direct substitution if the function is well-behaved at that point. But, when dealing with functions that involve a potential division by zero or other undefined expressions, such as in our exercise with the function \( \frac{x^2-1}{x-1} \) as \( x \) approaches 1, further simplification is often required beforehand, such as recognizing and factoring a difference of squares.
Difference of Squares
Recognizing patterns in algebra can be tremendously helpful, and the difference of squares is one of these critical patterns. The difference of squares is a special product form \( a^2 - b^2 \), which can be factored into \( (a+b)(a-b) \).

This algebraic identity is particularly useful when trying to simplify expressions or solve equations. It allows us to rewrite seemingly complex fractions or equations into simpler forms for easy evaluation or cancellation.

As demonstrated in the problem, the numerator \( x^2-1 \) is recognized as a difference of squares since it can be written as \( (x+1)(x-1) \). Factoring in this manner allows any common factors between the numerator and denominator to be omitted, which is a legitimate operation when seeking a limit—provided \( x \) does not exactly equal the value causing the zero denominator. Such simplification is critical for evaluating the limit of a function, which leads to determining the function's continuity at a particular point.
Continuity Checklist
Continuity at a point is a fundamental concept of calculus. To determine if a function is continuous at a specific point, we must satisfy three critical conditions, which can be considered a 'checklist' for continuity:

  • The function must be defined at the point.
  • The limit of the function as it approaches the point from both the left and right must exist.
  • The value of the function at the point must equal the limit as it approaches from both sides.

For the function in our exercise, although the first two conditions of the continuity checklist are satisfied—since the function \( f(x) \) is defined at \( x=1 \) and the limit exists as \( x \) approaches 1—the third condition is not satisfied because the function's value at \( x=1 \) does not match the limit value. Thus, by the continuity checklist, we can definitively conclude that the function is not continuous at \( x=1 \).

When solving problems related to continuity, always remember to run through this checklist. It will systematically guide you to a correct conclusion about a function's behavior at the targeted point.

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