Problem 16 - Calculus: Early Transcendentals

Question

Evaluate the limit, if it exists. \( \displaystyle \lim_{x \to -1}\frac{2x^2 + 3x + 1}{x^2 - 2x - 3} \)

Step-by-Step Solution

Answer

The limit is \(\frac{1}{4}\).

Step 1: Analyze Direct Substitution

First, we attempt direct substitution by substituting \(x = -1\) into the function:\[\frac{2(-1)^2 + 3(-1) + 1}{(-1)^2 - 2(-1) - 3} = \frac{2(1) - 3 + 1}{1 + 2 - 3} = \frac{0}{0}\]This results in an indeterminate form \(\frac{0}{0}\), indicating that we need to simplify the expression to resolve the limit.

Step 2: Factor the Numerator and Denominator

To simplify the expression and resolve the indeterminate form, factor both the numerator and the denominator.The numerator is \(2x^2 + 3x + 1\). It can be factored as:\[2x^2 + 3x + 1 = (2x + 1)(x + 1)\]The denominator is \(x^2 - 2x - 3\). It can be factored as:\[x^2 - 2x - 3 = (x + 1)(x - 3)\]

Step 3: Simplify the Expression

Now substitute the factored forms back into the original fraction:\[\frac{(2x + 1)(x + 1)}{(x + 1)(x - 3)}\]Cancel the common factor \((x + 1)\) from the numerator and the denominator:\[\frac{2x + 1}{x - 3}\]

Step 4: Evaluate the Simplified Limit

After canceling, we are left with the expression \(\frac{2x + 1}{x - 3}\). Now plug \(x = -1\) into this simplified expression:\[\frac{2(-1) + 1}{-1 - 3} = \frac{-2 + 1}{-4} = \frac{-1}{-4} = \frac{1}{4}\]

Step 5: Conclusion on the Limit

Since substituting \(x = -1\) into the simplified expression gives us the value \(\frac{1}{4}\), we conclude that the limit exists and is equal to \(\frac{1}{4}\).

Key Concepts

Indeterminate Forms

When evaluating limits, you might encounter the term **indeterminate forms**. An indeterminate form typically occurs when evaluating a limit through direct substitution results in outcomes like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms signify that the limit can't be directly determined, and further simplification is often needed.

In the problem at hand, substituting \(x = -1\) into \(\lim_{x \to -1}\frac{2x^2 + 3x + 1}{x^2 - 2x - 3}\) initially yields \(\frac{0}{0}\). This means the expression is indeterminate, and simply inserting \(x = -1\) doesn't provide the limit's value.

When encountering \(\frac{0}{0}\), additional algebraic techniques, like factoring or rationalizing, are necessary to alter the expression's form so a value can be determined.

Direct Substitution

**Direct substitution** is often the first step when evaluating limits. This method involves substituting the approaching value directly into the function to determine the limit.

In our example, substituting \(x = -1\) gives us an immediate \(\frac{0}{0}\) result, indicating an indeterminate form. However, direct substitution can sometimes provide a solution without additional work, especially if the function is continuous at the point in question.

When initial direct substitution fails due to an indeterminate form, we must alter the function, typically through factoring or simplifying, to recognize the limit.

Factoring Expressions

**Factoring expressions** helps unravel indeterminate forms by identifying and canceling common terms in the numerator and denominator.

In the function \(\frac{2x^2 + 3x + 1}{x^2 - 2x - 3}\), both the numerator and the denominator can be factored:

• The numerator \(2x^2 + 3x + 1\) becomes \((2x + 1)(x + 1)\).
• The denominator \(x^2 - 2x - 3\) simplifies to \((x + 1)(x - 3)\).

Factoring reveals a common factor, \((x + 1)\), in both parts. This commonality facilitates canceling it out to simplify the limit expression.

Ultimately, factoring transforms complex expressions into simpler forms that are more easily evaluated.

Simplifying Fractions

**Simplifying fractions** is crucial for resolving indeterminate forms and finding the actual limit.

Once you've factored both the numerator and the denominator like we did, the next step is cancelling out any common factors to simplify the expression:

• Original factored form: \(\frac{(2x + 1)(x + 1)}{(x + 1)(x - 3)}\)
• Cancel \((x + 1)\): \(\frac{2x + 1}{x - 3}\)

Now, direct substitution works without yielding an indeterminate form. Substituting \(x = -1\) into \(\frac{2x + 1}{x - 3}\) provides the limit as \(\frac{1}{4}\).

Simplifying until reaching a workable form makes it possible to accurately evaluate the limit.