Problem 24 - Precalculus

Question

Evaluate the following limits using direct substitution, if possible. If not possible, state why. $$\lim _{x \rightarrow-2} \frac{4-x^{2}}{x^{2}+5 x+6}$$

Step-by-Step Solution

Answer

Direct substitution results in \( \frac{0}{0} \), an indeterminate form, so it's not possible.

Step 1: Substitute the Limit Value Directly

To evaluate the limit \( \lim _{x \rightarrow -2} \frac{4-x^{2}}{x^{2}+5 x+6} \), we start by directly substituting \( x = -2 \) into the function. This gives us \( \frac{4 - (-2)^2}{(-2)^2 + 5(-2) + 6} \).

Step 2: Simplify the Numerator

Calculate \( 4 - (-2)^2 \). This becomes \( 4 - 4 = 0 \).

Step 3: Simplify the Denominator

Calculate \( (-2)^2 + 5(-2) + 6 \). This becomes \( 4 - 10 + 6 = 0 \).

Step 4: Analyze and Conclude

Since both the numerator and the denominator evaluate to 0, the expression is of the form \( \frac{0}{0} \), an indeterminate form. Thus, direct substitution is not possible, and further algebraic manipulation may be necessary to resolve the limit.

Key Concepts

Direct Substitution

In calculus, direct substitution is the first method to try when evaluating a limit. The idea is simple: you substitute the value that the variable is approaching directly into the function. For example, if you have a limit of a function as \( x \) approaches some number, you just plug that number into the function.
Here’s how it works:

• You take the limit expression \( \lim_{x \rightarrow a} f(x) \), where \( f(x) \) is your function and \( a \) is the value that \( x \) is approaching.
• Substitute \( a \) directly into the function, replacing \( x \) with \( a \).
• Simplify the expression, if possible, to get a number or determine that it's undefined.

Sometimes, direct substitution doesn’t work because you encounter problematic cases like the indeterminate form \( \frac{0}{0} \), which was the case in the original exercise. When both numerator and denominator converge to zero, it indicates that other methods might be needed to evaluate the limit. Nonetheless, direct substitution is usually the best first step since it is straightforward and often yields the result immediately.

Indeterminate Form

When substituting in a limit results in an expression like \( \frac{0}{0} \), you've encountered what's known as an indeterminate form. It's called indeterminate because you cannot determine the limit’s value directly from that outcome alone.
Indeterminate forms signal that there's more work to be done to evaluate the limit accurately.Here are a few things to keep in mind about indeterminate forms:

• \( \frac{0}{0} \) does not mean the limit doesn't exist, only that direct substitution wasn't sufficient.
• Other expressions, like \( \frac{\infty}{\infty} \) or \( \infty - \infty \), can also be indeterminate forms.
• These forms indicate the need for other strategies like algebraic manipulation, factoring, or applying L'Hôpital's Rule to find the limit.

Recognizing an indeterminate form is crucial as it directs you to seek alternative methods for finding the limit. Although it seems like a roadblock, it's often just a signpost pointing toward more advanced tools in calculus.

Algebraic Manipulation

After recognizing an indeterminate form, we often turn to algebraic manipulation to evaluate the limit. This involves rearranging, factoring, or simplifying the expression to eliminate the indeterminate form.In our example, by factoring the numerator and the denominator, we might be able to cancel out common factors, resolving the \( \frac{0}{0} \) issue.
To illustrate:

• Consider \( 4 - x^2 \) as \( (2-x)(2+x) \).
• The denominator \( x^2 + 5x + 6 \) can be factored into \( (x+2)(x+3) \).
• Both expressions share a common factor: \( x + 2 \). Cancelling these can lead to a new limit expression where direct substitution becomes applicable again.

Through this approach, algebra plays a pivotal role in transforming expressions to simplify the evaluation of limits. Mastering these techniques is vital, as they form the foundation for solving more complex calculus problems effectively.