Problem 55 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 1} \frac{x^{2}-1}{x+1} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Direct Substitution

Attempt to substitute the limit value directly into the expression. Substitute 1 for \( x \) in the function: \( \frac{(1)^2 - 1}{1 + 1} = \frac{0}{2} = 0 \).

Step 2: Simplify the Expression

To further simplify, factor the numerator. The expression becomes \( \lim_{x \rightarrow 1} \frac{(x-1)(x+1)}{x+1} \).

Step 3: Cancel Common Factors

Cancel the common term \( x+1 \) from both the numerator and the denominator. The expression simplifies to \( \lim_{x \rightarrow 1} (x-1) \).

Step 4: Substitute Again

Now substitute 1 into the simplified expression: \( x - 1 = 1 - 1 = 0 \).

Step 5: Final Calculation

The limit evaluates to 0, as directly substituting 1 into the simplified expression \( x - 1 \) yields 0.

Key Concepts

Direct substitution method

The Direct Substitution Method is a straightforward technique used to find limits. It involves plugging the value that x is approaching directly into the function. This is often the first step when evaluating limits because it can sometimes quickly provide the limit without further manipulation.
For example, if we have \[ \lim_{x \rightarrow 1} \frac{x^{2}-1}{x+1} \]and we directly substitute 1 in place of \( x \), we perform the operation:

• Calculate \( (1)^2 \) which equals 1.
• Then subtract 1 resulting in 0 for the numerator.
• The denominator becomes \( 1 + 1 = 2 \).

The fraction then becomes \( \frac{0}{2} \), which equals 0.
This first attempt shows that the expression is undefined or requires further steps, particularly if it leads to forms like \( \frac{0}{0} \). Thus, it signals the need for further simplification to find the correct limit.

Factoring polynomials

Factoring polynomials plays a crucial role in simplifying expressions for easier limit evaluation. This technique breaks down polynomials into simpler components that can make evaluating limits more manageable.
In our case, we took \[ \frac{x^{2} - 1}{x+1} \]and noticed it can be factored as:

• The numerator \( x^{2} - 1 \) is a difference of squares: \( x^{2} - 1 = (x-1)(x+1) \).
• Thus, the expression transforms to: \[ \lim_{x \rightarrow 1} \frac{(x-1)(x+1)}{x+1} \]

Factoring helps because it allows the expression to be broken down into parts, often revealing common factors that can be safely reduced, moving you closer to a simplified expression that’s easier to handle further.

Limit simplification

Limit simplification is often necessary when direct substitution leads to indeterminate forms like \( \frac{0}{0} \). After factoring the expression, we transition to simplifying the limit. The goal here is to create a simpler form that eliminates these obstacles.
For the exercise's expression \[ \lim_{x \rightarrow 1} \frac{(x-1)(x+1)}{x+1} \],factoring reveals possible simplifications. By recognizing the common factor \( (x+1) \) in both the numerator and the denominator, we prepare for the next step – canceling common factors. This simplification step is crucial to making the problem solvable through substitution.

Canceling common factors

Canceling common factors is the step where we simplify the expression by reducing the common elements in the numerator and the denominator. This is essential for making a limit problem like ours manageable, especially when it initially presents an indeterminate form.
In the expression \[ \frac{(x-1)(x+1)}{x+1} \],observe that \( (x+1) \) appears in both the numerator and denominator:

• Cancel out the \( (x+1) \), simplifying the expression to \( x-1 \).

This simplification removes the original problem of division by zero when \( x \) approaches 1. The limit then becomes:\[ \lim_{x \rightarrow 1} (x-1). \]We are left with a straightforward expression, allowing us to substitute directly once more to find the limit is indeed 0. This final step verifies the accuracy of our simplifications and substitutions, leading to the final solution.