Problem 21 - Calculus Early Transcendentals

Question

Use the precise definition of a limit to prove the following limits. $$\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}=8 \text { (Hint: Factor and simplify.) }$$

Step-by-Step Solution

Answer

Answer: The limit of the function \(\frac{x^{2}-16}{x-4}\) as x approaches 4 is 8.

Step 1: Rewrite the function using factoring

In order to simplify the given expression, we need to factor the numerator. Given a difference of squares \(\left(a^{2} - b^{2}\right)\), we can use the formula \(\left(a^{2} - b^{2}\right) = (a+b) (a-b)\). Here, the numerator is \(x^2 - 16\). We can rewrite it as \((x+4)(x-4)\) since \(a=x\) and \(b=4\). Now, the function becomes: $$\frac{(x+4)(x-4)}{x-4}$$

Step 2: Simplify the function

We can now simplify the function by canceling out the common terms in both the numerator and the denominator: $$\frac{(x+4)(x-4)}{x-4} = x+4$$ The function has been simplified to \(x + 4\).

Step 3: Apply the limit

Now that we have simplified the function, we can apply the limit as x approaches 4: $$\lim_{x \rightarrow 4} (x + 4) $$

Step 4: Prove the limit

Applying the limit of x approaches 4, we simply substitute 4 into the function: $$\lim_{x \rightarrow 4} (x + 4) = 4 + 4 = 8$$ We have shown that the given limit is equal to 8: $$\lim_{x \rightarrow 4} \frac{x^{2}-16}{x-4} = 8$$

Key Concepts

Factoring

Factoring is a crucial mathematical technique that simplifies complex expressions by breaking them down into products of simpler expressions. Imagine having a lump of clay that can be reshaped into smaller, manageable pieces—factoring does this with algebraic expressions.

In the context of our problem, we have the expression \(x^2 - 16\), which is a perfect example of the "difference of squares." This term stands for two squared numbers separated by a subtraction sign.

To factor such expressions, we use the formula \(a^2 - b^2 = (a + b)(a - b)\).

• In "\(x^2 - 16\)," \(x^2\) is \(a^2\) and \(16\) is \(b^2\), so \(a = x\) and \(b = 4\).
• Applying the formula gives us \((x + 4)(x - 4)\), transforming the original problem into simpler parts.

This technique helps us in further simplifying problems by making complicated polynomials easier to handle.

Difference of Squares

The difference of squares is a special case of factoring that often occurs in algebraic problems. It's one of the simplest and most easily recognizable formulas in algebra.

When we say "difference of squares," we're referring to a format where two squared terms are subtracted, like \(x^2 - 16\). The great thing about this pattern is the consistent way it factors:

• This expression \(x^2 - 16\) fits perfectly into the formula \(a^2 - b^2 = (a + b)(a - b)\), where the squares become apparent as \((x + 4)(x - 4)\).

This understanding allows us to not just factor expressions quickly but also simplify and solve equations more effectively.

Knowing these shortcuts can make algebra much more manageable and less intimidating.

Simplifying Expressions

After factoring an expression, simplifying it is the next logical step. Simplification involves reducing the expression to its most concise form by removing any unnecessary terms.

In our problem, after factoring \((x+4)(x-4)\), we encounter the same \((x-4)\) term in both the numerator and the denominator. This is a perfect opportunity for simplification:

• The \((x-4)\) terms cancel each other out, leaving us with just \(x+4\).

This cancels nearly all the complexity out of the original expression, turning it into something that's straightforward and much easier to evaluate. Simplification ensures that when we apply mathematical operations like limits, they become manageable.

Substitution Method

The substitution method is a fundamental technique for evaluating limits, particularly after simplifying an expression. This method involves replacing variables with specific values to compute a limit.

In this exercise, after simplifying the expression to \(x+4\), the task becomes straightforward. We simply substitute the value \(x=4\) into the expression:

• Replace, \(x\) with \(4\) in \(x+4\), getting \(4+4\) which equals \(8\).

This step confirms that the function approaches the value \(8\) as \(x\) reaches \(4\).

Substitution is a direct and often final step, validating the process by proving the outcome of a limit after prior work in factoring and simplification.