Problem 24 - Precalculus: Mathematics for Calculus

Question

Finding Limits Evaluate the limit if it exists. $$\lim _{x \rightarrow 2} \frac{x^{4}-16}{x-2}$$

Step-by-Step Solution

Answer

The limit is 32.

Step 1: Identify the Limit Form

First, determine the form of the expression as x approaches 2. Substitute 2 into the expression: \( \frac{2^4 - 16}{2 - 2} = \frac{0}{0} \). This tells us that the limit is in an indeterminate form \( \frac{0}{0} \), which means we need to simplify.

Step 2: Simplify the Expression

The expression \( x^4 - 16 \) can be factored. Notice that \( x^4 - 16 = (x^2 - 4)(x^2 + 4) \) because \( x^4 - 16 \) is a difference of squares. Furthermore, \( x^2 - 4 \) is also a difference of squares and can be factored into \( (x - 2)(x + 2) \). Therefore, \( x^4 - 16 = (x - 2)(x + 2)(x^2 + 4) \).

Step 3: Cancel Common Factors

The expression becomes \( \frac{(x - 2)(x + 2)(x^2 + 4)}{x - 2} \). Cancel the common factor of \( x - 2 \) from the numerator and the denominator: \( \frac{(x + 2)(x^2 + 4)}{1} = (x + 2)(x^2 + 4) \).

Step 4: Evaluate the Limit

Substitute \( x = 2 \) into the simplified expression \( (x + 2)(x^2 + 4) \). We find \( (2 + 2)((2)^2 + 4) = 4(4 + 4) = 4(8) = 32 \). Hence, \( \lim_{x \to 2} \frac{x^4 - 16}{x - 2} = 32 \).

Key Concepts

Evaluating Limits

Limits are fundamental in calculus and help us understand how a function behaves as it approaches a certain value. To evaluate limits, we replace the variable in the function with the point it is approaching. Sometimes, the limit can be determined by direct substitution. But in cases where substituting the value leads to an indeterminate form like \(\frac{0}{0}\), we need to simplify the expression first. Here's how to begin:

• Substitute the value into the function.
• Check if the result yields a determinate form or an indeterminate one.
• If indeterminate, attempt to simplify or rearrange the expression to solve the limit.

By using techniques like factoring or rationalizing, we simplify complex expressions and avoid indeterminate forms to accurately find the limit.

Difference of Squares

The difference of squares is a handy algebraic tool used to simplify expressions. It follows the formula \((a^2 - b^2) = (a + b)(a - b)\). In the given exercise, the numerator \(x^4 - 16\) fits the pattern since 16 is the square of 4. This transforms into \((x^2 - 4)(x^2 + 4)\). Notice that \(x^2 - 4\) is further factorable since it is also a difference of squares, \((x^2 - 4) = (x - 2)(x + 2)\). Using this approach helps break down algebraic expressions, making them easier to cancel terms when part of a fraction. This simplification often resolves indeterminate forms encountered in limit problems.

Indeterminate Forms

An indeterminate form happens when directly substituting a value into a function returns an undefined state, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). To resolve these forms, we apply algebraic modifications:

• Factoring: Simplify expressions by breaking them into products of simpler factors to cancel common terms.
• Rationalizing: Particularly useful when dealing with radicals, helping transform the expression into a determinate form.
• Limit Laws: Applying standard limit laws to simplify expressions.

In our exercise, resolving the \(\frac{0}{0}\) indeterminate form required factoring the difference of squares to eliminate the common term. Once simplified, substituting the limit point yields a tangible result, like the 32 obtained in the example, making the solution clear and accurate.