Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 49-62, find the indicated limit, if it exists. \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}\)

Step-by-Step Solution

Answer

The simplified expression after canceling out the common factor is \(x+1\). Therefore, the limit as x approaches 1 is \(\lim_{x\rightarrow1} (x+1) = 2\).

Step 1: Identify the Limit to Evaluate

The exercise asks us to find the limit of the given function as x approaches 1: \[ \lim_{x\rightarrow1} \frac{x^2-1}{x-1} \]

Step 2: Simplify the Expression

Notice that the numerator of the function is a difference of squares, which can be factored as follows: \[ x^2-1=(x-1)(x+1) \] Now, substitute this factored form into the expression: \[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} \] We can now cancel the common factor in the numerator and the denominator: \[ \frac{(x-1)(x+1)}{x-1} = x+1 \]

Step 3: Evaluate the Limit Directly

Now that we have simplified the expression, we can directly evaluate the limit: \[ \lim_{x\rightarrow1} (x+1) \] Substitute the value x=1 into the simplified expression: \[ (1)+1=2 \] Hence, the limit is: \[ \lim_{x\rightarrow1} \frac{x^2-1}{x-1} = 2 \]

Key Concepts

Limits

Limits are a fundamental concept in calculus that describe how a function behaves as its input approaches a certain value. Understanding limits is essential because they form the backbone of derivative and integral calculations, which are central to calculus.

When we say \(\lim_{x \to a} f(x) = L\), it means that as \(x\) gets infinitely close to \(a\), the function \(f(x)\) approaches the value \(L\). However, the function doesn't necessarily need to equal \(L\) at \(x = a\). Limits help us describe the behavior of functions when direct evaluation is tricky or impossible.

• The limit might exist if approaches a specific number.
• Sometimes, a limit doesn't exist due to function behavior, such as discontinuities or vertical asymptotes.
• When solving limits, especially with indeterminate forms like \(\frac{0}{0}\), algebraic manipulation is often necessary.

Factoring

Factoring is a technique used to break down algebraic expressions into simpler components, which can help simplify math problems, including limits. It involves expressing a polynomial as a product of its factors.

One of the most common factoring techniques is identifying special patterns, such as the difference of squares. Factoring not only simplifies calculations but also allows us to cancel common terms, leading to easier evaluations.

• The expression \(x^2 - 1\) can be factored into \((x - 1)(x + 1)\).
• This step allows us to remove or cancel common factors in fractional expressions.
• Factoring is crucial when dealing with polynomial limits to avoid zero denominators.

After factoring, challenges like dividing by zero in our limits become manageable by canceling equivalent terms in the numerator and denominator.

Difference of Squares

The difference of squares is a specific algebraic identity that applies to expressions of the form \(a^2 - b^2\), and it can be factored into \((a - b)(a + b)\). Recognizing this pattern can simplify complex expressions and is particularly helpful when solving limits.

In the expression \(x^2 - 1\), \(1\) can be viewed as \(1^2\), making it a classic difference of squares:

• Rewrite \(x^2 - 1^2\) as \((x - 1)(x + 1)\).
• Use this factorization to simplify limits or solve equations with similar forms.
• Understanding and applying this identity avoids common pitfalls in limit calculations.

The difference of squares is a key tool, often used in simplifying functions to make the calculation of limits more straightforward.

Evaluating Limits

Evaluating limits involves the process of finding the value a function approaches as the input approaches a specific point. There are several direct and indirect methods for determining limits, depending on the function’s form and behavior.

In our initial step-by-step solution:

• We started by factoring to handle the "0/0" indeterminate form.
• Next, we canceled the common factor \(x - 1\) to simplify the expression.
• The final step allowed us to substitute \(x = 1\) into the simplified expression \(x + 1\), resulting in 2.

Direct substitution often works after simplification. If substitution initially gives an indeterminate form, like \(\frac{0}{0}\), methods such as factoring, rationalizing, or applying L'Hôpital's Rule might be needed.

Evaluating limits is about transforming the function to a state where its behavior near the target value is clear, ensuring problems are analytically solved.