Problem 367

Question

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

Verified
Answer
The limit is 6.
1Step 1: Determine the Form of the Limit
First, substitute \(x = 3\) directly into the expression \(\frac{x^2 - 9}{x - 3}\) to determine its form. \[ \text{Substitute } x = 3: \frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0} \] Since the expression results in an indeterminate form \(\frac{0}{0}\), we can use alternative algebraic methods or L'Hôpital's Rule to find the limit.
2Step 2: Simplify the Expression by Factoring
Notice that the numerator \(x^2 - 9\) is a difference of squares and can be factored: \[x^2 - 9 = (x - 3)(x + 3)\] Thus, the expression becomes: \[\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3}\] Cancel \(x - 3\) from the numerator and denominator, assuming \(x eq 3\).
3Step 3: Solving for the Limit After Simplification
After canceling \(x - 3\), the expression simplifies to \(x + 3\). Now, substitute \(x = 3\) into the simplified expression: \[ x + 3 = 3 + 3 = 6 \] Thus, the limit is 6.

Key Concepts

L'Hôpital's RuleIndeterminate FormsFactoring Polynomials
L'Hôpital's Rule
L'Hôpital's Rule is a handy tool in calculus, especially when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule offers a method to evaluate limits that are difficult to solve using algebraic simplification or direct substitution.
To apply L'Hôpital's Rule, the function you are evaluating must fit certain criteria:
  • Both the numerator and the denominator approach 0 or \(\infty\) as \(x\) approaches the limit.
  • The derivatives of the numerator and denominator exist and are continuous near the point of interest.
When these conditions are met, you can differentiate the numerator and the denominator separately, then take the limit of their ratio:\[\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}\]Applying L'Hôpital's Rule may sometimes reveal different forms or even simplify the expression further.
It's crucial to verify that conditions hold after applying the rule, ensuring both derivatives are properly calculated and are not resulting in additional indeterminate forms.
Indeterminate Forms
In calculus, an indeterminate form is an expression whose limit cannot be directly determined from its initial setup. Some common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\infty - \infty\), and product or power forms like \(0 \cdot \infty\) and \(1^\infty\).
Encountering these forms signifies uncertainty, warranting additional techniques such as algebraic manipulation or applying L'Hôpital's Rule to find a solvable expression.
For instance, in the original exercise, substituting \(x = 3\) resulted in \(\frac{0}{0}\), showcasing an indeterminate form. Practical handling involves designating a suitable approach — such as simplifying before applying a limit — maintaining the integrity of the expression while uncovering its true limit.
Being comfortable with indeterminate forms is crucial as they frequently appear in calculus problems and understanding them aids in choosing the right problem-solving methods.
Factoring Polynomials
Factoring polynomials is a straightforward technique often used to simplify algebraic expressions, making limits easier to evaluate. Polynomials like \(x^2 - 9\) are sometimes perfect squares or other recognizable forms.
The expression \(x^2 - 9\) in the exercise is a classic difference of squares, easily factored into \((x - 3)(x + 3)\). This transforms the problematic term into one that can often be simplified further.
In our exercise, factoring allowed for the cancellation of \(x - 3\) in the numerator and the denominator, revealing a simpler expression. Post-cancellation, direct substitution becomes viable again for limit evaluation, offering a neat solution path to the previously indeterminate form.
Understanding different factoring techniques for polynomials aids in resolving many algebraic issues and is an essential skill across various levels of mathematics, particularly in calculus limits evaluation.