Problem 2 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 2} \frac{x^{2}+x-6}{x-2}$$

Step-by-Step Solution

Answer

The limit is 5.

Step 1: Evaluate the Limit Directly

First, substitute the value of the limit into the expression: \[\frac{x^2 + x - 6}{x - 2} \quad \text{at} \quad x = 2\]Substituting, gives:\[\frac{2^2 + 2 - 6}{2 - 2} = \frac{4 + 2 - 6}{0} = \frac{0}{0}\]Since this results is an indeterminate form, we need to find another approach.

Step 2: Factor the Expression

Factor the numerator of the expression to simplify it:\[x^2 + x - 6 = (x - 2)(x + 3)\]This simplifies the original limit to:\[\frac{(x-2)(x+3)}{x-2}\]

Step 3: Cancel Common Factors

Notice that there is a common factor of \(x-2\) in both the numerator and the denominator. Cancel these common terms:\[\frac{(x-2)(x+3)}{x-2} = x + 3\]

Step 4: Evaluate the Simplified Limit

Now substitute \(x = 2\) into the simplified expression:\[x + 3 \quad \text{at} \quad x = 2\]Thus,\[2 + 3 = 5\]

Step 5: Conclusion

The limit of the original expression as \(x\) approaches 2 is 5. We used factorization instead of l'Hospital's Rule since it was more straightforward.

Key Concepts

Understanding Indeterminate Forms

When dealing with limits, you may encounter expressions where direct substitution results in forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are known as "indeterminate forms". Indeterminate forms are called so because they do not convey useful information about the limit's value.
If not handled, they make it difficult to analyze and solve limit problems directly.
Handling indeterminate forms appropriately often requires algebraic manipulation or rules like l'Hospital's Rule.
For example, in the given exercise, substituting \(x = 2\) gives \(\frac{0}{0}\), a classic indeterminate form.
This indicates the need for strategies like factorization or applying l'Hospital's Rule to simplify and evaluate the limit.

The Power of Factorization

Factorization is a powerful tool that helps simplify algebraic expressions, especially when dealing with limits.
By factoring polynomials, you can often cancel out problematic terms that lead to an indeterminate form.
In the given exercise, factorization was used effectively to simplify the expression \(x^2 + x - 6\).
Factorization involves finding two expressions that, when multiplied, produce the original expression.
Here, \(x^2 + x - 6\) is factored into \((x - 2)(x + 3)\).
This factorized form reveals a common term in the numerator and denominator, \(x - 2\).
Canceling the \((x - 2)\) terms removes the indeterminate form, simplifying the problem and making it easier to evaluate the limit.

Grasping Limits in Calculus

Limits play a fundamental role in calculus, providing the foundation for concepts like continuity, derivatives, and integrals.
They describe the behavior of a function as it approaches a specific point or extends towards infinity.
In practical terms, calculating a limit involves determining how a function behaves near a particular value.
In the original exercise, we aim to find the limit as \(x\) approaches 2 for the function \(\frac{x^2+x-6}{x-2}\).
This process often involves techniques like substitution, simplification, and sometimes using rules such as l'Hospital's Rule.
Understanding limits is crucial for mastering other calculus concepts, as they often rely on the behavior of a function near certain points or values.