Problem 33 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$$

Step-by-Step Solution

Answer

Answer: The limit is 2.

Step 1: Factor the numerator using the difference of squares

The numerator of the given expression is \(x^2 - 1\). We can factor it using the difference of squares formula \((a^2-b^2) = (a+b)(a-b)\). Therefore, \(x^2 - 1\) can be factored as \((x+1)(x-1)\). So, the expression becomes: $$\frac{(x+1)(x-1)}{x-1}$$

Step 2: Simplify the expression by canceling out the common term

Since \((x-1)\) is a common term in both the numerator and the denominator, we can cancel it out. The expression now becomes: $$x+1$$

Step 3: Evaluate the limit

Now that the expression has been simplified, we can find the limit as x approaches 1. $$\lim _{x \rightarrow 1} (x+1)$$ Substitute 1 for x: $$1+1=2$$ Therefore, the limit is 2. So, $$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} = 2$$.

Key Concepts

Difference of Squares

Understanding the difference of squares is critical when you're trying to factor polynomials and can be incredibly helpful in simplifying expressions to evaluate limits in calculus. This concept revolves around the fact that a binomial of the form \(a^2 - b^2\) can be broken down into the product of two simpler binomials, \(a + b\) and \(a - b\). It’s like looking at a square of side 'a' and subtracting a smaller square of side 'b' to see the space that remains, which is the difference between their areas.

For example, if you have the expression \(x^2 - 1\), you can identify 'x' as 'a' and '1' as 'b'. Applying the difference of squares rule gives you \(x + 1\) and \(x - 1\) which multiplies back to the original expression \(x^2 - 1\). This technique is a cornerstone in simplifying many algebraic expressions, especially when evaluating limits where you need to overcome indeterminate forms.

Factoring Polynomials

What Is Factoring?

Factoring polynomials is like breaking down a composite number into its prime factors—only here, we deal with algebraic expressions. The greatest common factor (GCF), grouping, and rules like the difference of squares are all methods used to factor polynomials into simpler, multipliable parts.

Importance in Limits

In the context of limits, factoring can be extremely useful in simplifying complex expressions, which can then lead to easily finding the limit value. Factoring helps eliminate common terms in the numerator and denominator, resolving indeterminate forms such as \(\frac{0}{0}\) which often appear when trying to directly substitute the limit value into a function.

Simplifying Expressions

Simplifying expressions is a crucial skill in calculus, as it can transform a seemingly complicated problem into a more manageable one. This process includes combining like terms, factoring, and reducing fractions to their simplest form. When you simplify an expression before evaluating a limit, you might eliminate terms that cause the function to be undefined at certain points.

For example, in the case of our exercise, simplifying the expression \(\frac{x^{2}-1}{x-1}\) by factoring and canceling out the common term \(x-1\) gives us \(x+1\), a simpler expression that can be directly evaluated as x approaches 1. It streamlines the computation and clarifies the behavior of the function around the point of interest.

Evaluating Limits

Approaching the Point

Evaluating limits involves determining what value a function approaches as the input (typically \(x\)) gets arbitrarily close to a certain point. It's a fundamental concept in calculus because it provides insight into the function's behavior without requiring the function to be actually defined at that point.

L’Hôpital’s Rule and Continuity

There are techniques such as L’Hôpital's Rule that help evaluate limits that result in indeterminate forms. Additionally, knowing whether a function is continuous at the point in question can simplify limit evaluation—the limit value is simply the function's value at that point. Our exercise is a straightforward example where, after simplification, the limit could be found by direct substitution since the simplified function is continuous at \(x = 1\).