Problem 24
Question
Determine each limit, if it exists. $$\lim _{x \rightarrow-2} \frac{x^{2}-4}{x+2}$$
Step-by-Step Solution
Verified Answer
The limit is -4.
1Step 1: 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.
2Step 2: 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 \(\frac{(x-2)(x+2)}{x+2}\).
3Step 3: 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.
4Step 4: 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 \(-4\).
Key Concepts
Indeterminate FormsFactoringSimplification
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.
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:
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.
- The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\).
- For \(x^2 - 4\), this becomes \((x-2)(x+2)\).
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:
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.
Remember:
- Canceling is only valid if the terms are not zero.
- In this scenario, do not reintroduce the canceled term as zero.
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.
Other exercises in this chapter
Problem 24
By considering the graph of the function but not calculating any limits, give the value of \(f^{\prime}(2)\) for each function. $$f(x)=x$$
View solution Problem 24
Use a table and/or graph to find the asymptote\((s)\) of each function. $$f(x)=\frac{e^{x}}{e^{x}-1}$$
View solution Problem 24
Complete each table and use the results to predict the indicated limit, if it exists. $$\text { If } f(x)=\frac{\sqrt{x}-3}{x-3}, \text { find } \lim _{x \right
View solution Problem 25
By considering the graph of the function but not calculating any limits, give the value of \(f^{\prime}(2)\) for each function. $$f(x)=-x$$
View solution