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Chapter 13: Problem 24

Determine each limit, if it exists. $$\lim _{x \rightarrow-2} \frac{x^{2}-4}{x+2}$$

Short Answer

Expert verified
The limit is -4.

Step by step solution

01

Substitute x with -2

To determine if the limit exists, we first substitute \(x = -2\) directly into the expression \(\frac{x^2-4}{x+2}\). This gives us \(\frac{(-2)^2-4}{-2+2}\), which results in \(\frac{4-4}{0}\), or \(\frac{0}{0}\). This is an indeterminate form, so we need to simplify further.
02

Factor the Numerator

Given the indeterminate form, we need to simplify \(\frac{x^2-4}{x+2}\). Notice that the numerator \(x^2-4\) is a difference of squares and can be factored as \( (x-2)(x+2) \). So, the expression becomes \(\frac{(x-2)(x+2)}{x+2}\).
03

Simplify the Expression

Next, we simplify \(\frac{(x-2)(x+2)}{x+2}\) by canceling out the common factor \(x+2\). This simplifies the expression to \(x-2\), as long as \(xeq-2\) because division by zero is undefined.
04

Evaluate the Simplified Expression

Now that we have the expression \(x-2\), we can substitute \(x = -2\) again. This gives us \(-2 - 2 = -4\). Therefore, the limit is \(-4\).

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Key Concepts

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

Indeterminate Forms
When working with limits, encountering an indeterminate form such as \(\frac{0}{0}\) is quite common. This form tells us that substituting directly doesn't give a clear answer for the limit. Thus, when calculating \(\lim _{x \rightarrow -2} \frac{x^{2}-4}{x+2}\), direct substitution resulted in \(\frac{0}{0}\).
This is a classic indeterminate scenario because the zero in the denominator suggests infinite values, while the zero in the numerator suggests convergence to zero, making it impossible to determine the limit without further manipulation.
Hence, finding the exact limit requires transforming the expression to resolve this ambiguous form.
Factoring
Factoring is a crucial algebraic technique used to simplify expressions, especially when dealing with indeterminate forms like \(\frac{0}{0}\). In the given limit problem, \(x^2 - 4\) can be rewritten using the difference of squares formula:
  • The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\).
  • For \(x^2 - 4\), this becomes \((x-2)(x+2)\).
By factoring the numerator, the original limit expression \(\frac{x^2-4}{x+2}\) transforms into \(\frac{(x-2)(x+2)}{x+2}\).
This step is vital as it allows us to identify and eliminate factors that might be causing the indeterminate form, specifically resolving the division by zero issue.
Simplification
Once the numerator is factored, simplification allows us to cancel out common terms. With \(\frac{(x-2)(x+2)}{x+2}\), the \(x+2\) in the numerator and denominator cancels, simplifying the expression to \(x - 2\).
Remember:
  • Canceling is only valid if the terms are not zero.
  • In this scenario, do not reintroduce the canceled term as zero.
Now, substitution can occur without issues. Evaluating \(x - 2\) at \(x = -2\) gives \(-2 - 2 = -4\).
This straightforward simplification highlights how overcoming an indeterminate form and leveraging algebraic techniques such as factoring can lead to determining the exact value of a limit.

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Most popular questions from this chapter

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