Problem 368 - Calculus Volume 1

Question

For the following exercises, evaluate the limits with either L'Hôpital's rule or previously learned methods $$ \lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3} $$.

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understanding the Limit Setup

The given limit is \( \lim _{x \rightarrow 3} \frac{x^{2}-9}{x+3} \). First, substitute \( x = 3 \) into the expression to check if it results in an indeterminate form. We get \( \frac{3^2 - 9}{3 + 3} = \frac{0}{6} = 0 \). This is not an indeterminate form, so direct substitution gives the limit value without further steps.

Step 2: Attempt direct substitution

Try substituting the value the variable approaches directly into the expression.

Step 3: Handle indeterminate forms

If direct substitution gives an indeterminate form, apply L'Hopital's Rule, algebraic manipulation, or other techniques.

Step 4: Evaluate the limit

Compute the final value of the limit.

Step 5: State the conclusion

Express the final answer.

Key Concepts

L'Hôpital's Rule

When evaluating limits, there are times when substitution results in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In these cases, L'Hôpital's Rule can be a powerful tool. This rule allows you to find the limit of a quotient by differentiating the numerator and denominator separately and then taking the limit of that new quotient.
To apply L'Hôpital's Rule effectively, you must first confirm the expression is indeed an indeterminate form. Once confirmed, differentiate both the top and bottom, but be careful to verify the new limit is determinate after substitution. Sometimes, it might need more than one application of the rule if the result is still indeterminate. Remember:

• L'Hôpital's Rule only applies to indeterminate forms \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).
• After differentiating, always check the result by substituting again.
• In cases where derivatives don't resolve the indeterminacy, consider other methods.

Direct Substitution

Direct substitution is one of the simplest and most straightforward methods for evaluating limits. If by substituting the value into the function, it doesn't result in an indeterminate form, you can directly calculate the result. This method is particularly convenient because it requires minimal computation.
For the given exercise, substituting \(x = 3\) into \( \frac{x^{2}-9}{x+3} \) quickly simplifies the expression, leading to a non-indeterminate form \( \frac{0}{6} = 0 \). This was a clear case where L'Hôpital's Rule wasn't necessary.
Total reliance on substitution depends on ensuring the function remains defined around the given point. If direct substitution results in division by zero or other undefined behaviors, other strategies like algebraic manipulation or L'Hôpital's Rule might be needed.

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating limits yields an expression like \( \frac{0}{0} \), \( \infty - \infty \), \( 0 \times \infty \), or other ambiguous forms. These forms do not directly provide enough information to determine the limit, thus requiring further analysis.
Recognizing indeterminate forms is crucial for deciding which limits techniques to apply. Generally, such forms demand alternative strategies, like algebraic simplification or L'Hôpital's Rule, to evaluate the limit properly.
Here are common indeterminate forms in limits:

• \( \frac{0}{0} \) - Often solved using L'Hôpital's Rule or factoring.
• \( \frac{\infty}{\infty} \) - Also typically resolved with L'Hôpital's Rule.
• \( \infty - \infty \) - Requires algebraic adjustments for resolution.
• \( 0 \times \infty \) - Re-expressed by converting to a fraction.
• \( 1^\infty, 0^0, \text{or} \infty^0 \) - These forms often need a logarithmic transformation.

In the given problem, substituting \(x = 3\) didn't result in an indeterminate form, allowing for straightforward evaluation. Understanding when an expression is indeterminate simplifies the process significantly.