Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

Many software packages have programs for calculating limits, although you should be warned that they are not infallible. To develop confidence in your program, use it to recalculate some of the limits in Problems 1-28. Then for each of the following, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{2 x+2}-2} $$

Step-by-Step Solution

Answer

The limit is 6.

Step 1: Substitute the limit into the function

First, try directly substituting the value as \( x \) approaches 1. Substitute \( x = 1 \) into the expression: \( \frac{1^3 - 1}{\sqrt{2 \cdot 1 + 2} - 2} \). This results in \( \frac{0}{0} \), which is an indeterminate form.

Step 2: Simplify the numerator

Notice the numerator \( x^3 - 1 \) can be factored using the difference of cubes formula: \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). So, factor \( x^3 - 1 \) as \( (x-1)(x^2 + x + 1) \).

Step 3: Simplify the denominator

Inspect the denominator \( \sqrt{2x+2} - 2 \). To simplify, multiply the expression by its conjugate \( \sqrt{2x+2} + 2 \) over itself to eliminate the square root: \( \frac{\sqrt{2x+2} - 2}{\sqrt{2x+2} - 2} \times \frac{\sqrt{2x+2} + 2}{\sqrt{2x+2} + 2} = \frac{(\sqrt{2x+2})^2 - 2^2}{(\sqrt{2x+2} + 2)(\sqrt{2x+2} - 2)} = \frac{2x + 2 - 4}{\sqrt{2x+2} + 2} \). This simplifies to \( \frac{2x - 2}{\sqrt{2x+2} + 2} = \frac{2(x-1)}{\sqrt{2x+2} + 2} \).

Step 4: Substitute and cancel terms

The expression \( \frac{x^3 - 1}{\sqrt{2x+2} - 2} \) is now \( \frac{(x-1)(x^2 + x + 1)}{\frac{2(x-1)}{\sqrt{2x+2} + 2}} \). Cancel the common factor \( (x-1) \) in both the numerator and denominator. This leaves \( \frac{x^2 + x + 1}{\frac{2}{\sqrt{2x+2} + 2}} = (x^2 + x + 1) \cdot \frac{\sqrt{2x+2} + 2}{2} \).

Step 5: Take the limit as \( x \rightarrow 1 \)

Now that the expression is no longer indeterminate, substitute \( x = 1 \) into \( (x^2 + x + 1) \cdot \frac{\sqrt{2x+2} + 2}{2} \). This results in \( (1^2 + 1 + 1) \cdot \frac{\sqrt{2 \cdot 1 + 2} + 2}{2} = 3 \cdot \frac{4}{2} = 3 \cdot 2 = 6 \).

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form arises when a mathematical expression leads to an unclear or undefined result after substitution. A common form is \( \frac{0}{0} \), which occurs when both the numerator and the denominator of a fraction approach zero simultaneously. This doesn't provide enough information about the limit, indicating that the function needs further simplification.

The goal is to manipulate the expression, usually through algebraic techniques such as factoring or rationalizing, to remove the indeterminate state. Once simplified, you can directly substitute the limit value to find a clear result.

Identifying and resolving indeterminate forms is an essential skill in calculus, as it allows for the evaluation of limits that initially appear confusing or unsolvable.

Difference of Cubes

The term 'difference of cubes' refers to an algebraic expression structured as \( a^3 - b^3 \). This can be factored into \((a-b)(a^2 + ab + b^2)\). This formula is vital for breaking down complex expressions into simpler, more manageable parts. In our original problem, recognizing \( x^3 - 1 \) as a difference of cubes is key to simplifying the expression and resolving the indeterminate form.

By substituting \(a = x\) and \(b = 1\), we could write \(x^3 - 1\) as \((x-1)(x^2 + x + 1)\). This places an \(x-1\) term in the numerator, crucial for cancellation later on. The difference of cubes is a powerful tool that should be well-understood, as it frequently appears in limit problems and other calculations.

Conjugate Multiplication

Conjugate multiplication is a method used to eliminate radicals or complex numbers in the denominator of an expression. This technique involves multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate is formed by changing the sign between two terms.

In our exercise, we dealt with \( \sqrt{2x+2} - 2 \). Its conjugate is \( \sqrt{2x+2} + 2 \). By multiplying by \( \frac{\sqrt{2x+2} + 2}{\sqrt{2x+2} + 2} \), we effectively rationalize the denominator, doing away with the square root.

This simplification step is often crucial to transforming an expression into a form where limits can be calculated directly, as it uncovers hidden factors that can be cancelled, resolving the indeterminate form.

Limit Simplification

Limit simplification is the process of reducing a complex limit expression to a form where it can be easily evaluated. It often involves algebraic manipulation, factoring, or using conjugates to expose and cancel out terms.

After simplifying the indeterminate expression \( \frac{(x-1)(x^2 + x + 1)}{\frac{2(x-1)}{\sqrt{2x+2} + 2}} \), the \( (x-1) \) terms cancel, significantly simplifying the process of finding the limit. This allows us to substitute and evaluate the limit directly, leading to a clear result.

The key to limit simplification is recognizing patterns and structures within expressions. Being able to deconstruct and rationalize these expressions helps solve limits that initially seem unsolvable.