Problem 30 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow-1} \frac{(x+1)^{2}}{2 x^{2}-x-3}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Factor the denominator

\(2x^2 - x - 3 = (2x - 3)(x + 1)\)

Verification: \((2x-3)(x+1) = 2x^2 + 2x - 3x - 3 = 2x^2 - x - 3\) \(\checkmark\)

Step 2: Simplify the expression

\(\frac{(x+1)^2}{2x^2 - x - 3} = \frac{(x+1)^2}{(2x-3)(x+1)} = \frac{x+1}{2x-3}\)

This simplification is valid for \(x \neq -1\).

Step 3: Evaluate the limit

\(\lim_{x \to -1} \frac{x+1}{2x-3} = \frac{-1+1}{2(-1)-3} = \frac{0}{-5} = \boxed{0}\)

Key Concepts

Direct Substitution

In calculus, one of the simplest methods to find the limit of a function as it approaches a certain point is through **direct substitution**. This involves taking the value that the variable is approaching—often called the limit point—and substituting it directly into the function.

**Simple Steps for Direct Substitution**

• Identify the value that the variable approaches in the limit (in our example, this is \(x = -1\)).
• Substitute this value into the function.
• Simplify the resulting expression to find the limit.

In the given exercise, substituting \(x = -1\) into the function's expression \(\frac{(x+1)^2}{2x^2-x-3}\) allowed us to directly calculate the limit without any additional steps since the operation yielded a valid result.

If after substituting, the function gives a form like \(\frac{0}{0}\), this indicates an indeterminate form, and other methods beyond direct substitution must be used. However, when the substitution results in a finite limit, as in this instance, direct substitution is both efficient and sufficient.

Continuity

**Continuity** is a fundamental concept when discussing limits, and it simplifies the process of finding them significantly. A function is continuous at a point if the following conditions are met:

• The function is defined at that point.
• The limit of the function as it approaches the point exists.
• The value of the function at that point equals the limit.

For the function \(f(x) = \frac{(x+1)^2}{2x^2-x-3}\), if a function is continuous at \(x = -1\), calculating the limit as \(x\) approaches \(-1\) by direct substitution is straightforward. Continuity tells us there will be no sudden jumps or holes in the graph of the function, hence making the direct substitution approach feasible.

It is important to note that if the denominator of a function becomes zero at a certain point, the function is not continuous at that point, leading potentially to undefined or infinite limits. This is why checking continuity around the limit point is crucial.

Numerator and Denominator Analysis

**Numerator and Denominator Analysis** plays a critical role in understanding the behavior of limits, particularly in rational functions like the one in the problem. Analyzing the numerator and the denominator separately helps determine the value of the function as it approaches a certain point.

**Key Points to Consider:**

• If the numerator equals zero and the denominator does not equal zero upon substitution, the limit is zero. This follows because any numerator of zero with a non-zero denominator results in zero.
• If both the numerator and denominator equal zero, it is termed an "indeterminate form," and requires methods like factoring or L'Hôpital's Rule to solve.
• If the denominator equals zero while the numerator does not, the limit tends generally towards infinity or does not exist.

For the given exercise, evaluating \(\frac{(x+1)^2}{2x^2-x-3}\) at \(x = -1\) results in \(\frac{0}{-2}\). The non-zero denominator confirms the limit to be zero. Analyzing both parts of the fraction allows us to conclude with certainty about the limit's behavior and existence.