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Chapter 1: Problem 17

Discuss the continuity of each function. $$ f(x)=\frac{x^{2}-1}{x+1} $$

Short Answer

Expert verified
The function \( f(x)=\frac{x^{2}-1}{x+1} \) is continuous for all \( x \) except at \( x = -1 \).

Step by step solution

01

- Simplify the function

First, simplify the function if possible. The function \( f(x) = \frac{x^{2}-1}{x+1} \) can be simplified by factoring the numerator. It becomes \( f(x) = \frac{(x-1)(x+1)}{x+1} \). This simplifies to \( f(x) = x-1 \), but this is valid only when \( x \neq -1 \) because at \( x = -1 \), the original function is undefined.
02

- Check the limit at x = -1

Next, check if the limit of the function as \( x \) approaches -1 exists. We find \( \lim_{{x \to -1^-}}f(x) = \lim_{{x \to -1^+}}f(x) = -2 \).
03

- Analyze the results

With the function simplified to \( f(x) = x-1 \) except at x = -1, we find that the function is continuous for all \( x \neq -1 \). For \( x = -1 \), although the limit exists and is -2, the function is not defined at \( x = -1 \). Hence the function is not continuous at \( x = -1 \).

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