Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

, 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: Substitution Check

First, try to substitute the value of 1 directly to check if the limit evaluates to a determinate form. Substitute 1 for \(x\) in the expression \(\frac{x^3 - 1}{\sqrt{2x+2} - 2}\). You obtain \(\frac{1^3 - 1}{\sqrt{2 \cdot 1 + 2} - 2} = \frac{0}{0}\), which is an indeterminate form.

Step 2: Simplifying the Numerator

Notice \(x^3 - 1\) can be rewritten as a difference of cubes: \(x^3 - 1 = (x-1)(x^2 + x +1)\). This will help in canceling out the indeterminate form.

Step 3: Simplifying the Denominator

To further simplify, multiply the numerator and the denominator by the conjugate of the denominator: \(\sqrt{2x+2} + 2\). The expression becomes: \[\frac{(x-1)(x^2 + x + 1)(\sqrt{2x+2} + 2)}{(\sqrt{2x+2} - 2)(\sqrt{2x+2} + 2)} = \frac{(x-1)(x^2 + x + 1)(\sqrt{2x+2} + 2)}{2x + 2 - 4}\].This results in: \[\frac{(x-1)(x^2 + x + 1)(\sqrt{2x+2} + 2)}{2x - 2}\].

Step 4: Cancel Common Factors

Notice that the denominator can also be factored: \(2(x-1)\). Cancel out the common factor \((x-1)\) from the numerator and denominator. This simplifies to \[\frac{(x^2 + x + 1)(\sqrt{2x+2} + 2)}{2}\].

Step 5: Evaluate the Limit

Substitute \(x = 1\) into the simplified expression: \[\frac{((1)^2 + (1) + 1)(\sqrt{2 \cdot 1+2} + 2)}{2} = \frac{3 \cdot (\sqrt{4} + 2)}{2} = \frac{3 \cdot (2 + 2)}{2} = \frac{3 \cdot 4}{2} = 6\].The limit is evaluated to 6, which means the limit exists.

Key Concepts

L'Hôpital's Rule

When facing difficult limit problems, especially when direct substitution leads to an indeterminate form like \( \frac{0}{0} \), L'Hôpital's Rule can rescue you. This rule helps us in evaluating limits by using derivatives. Essentially, if you have an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the rule allows you to differentiate the numerator and the denominator separately and then reevaluate the limit. This means:

• If \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\infty}{\infty} \),
• Then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), \( \text{given the new limit exists.} \)

In the given exercise, L'Hôpital's Rule wasn't used directly; instead, different algebraic manipulation techniques were applied to resolve the indeterminate form.

Indeterminate Forms

Indeterminate forms are a concept you come across when exploring limits in calculus. They occur when an expression initially leads to a form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) when you substitute a particular value into a function. These do not immediately determine the limit, hence the term 'indeterminate'.

• Typical indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \).
• In our example, substituting \( x = 1 \) led to a \( \frac{0}{0} \) form. This signaled the need for further steps to evaluate the limit properly through algebraic manipulation.

Rationalization

Rationalization is a handy technique used when dealing with limits that have expressions with square roots. The idea here is to multiply the numerator and the denominator by the conjugate of the denominator.

• The conjugate of a term like \( \sqrt{a} - b \) is \( \sqrt{a} + b \).
• This technique helps eliminate the square root in the denominator, simplifying the expression into something more manageable.

In the given exercise, rationalization helped simplify the original problem. By multiplying by the conjugate \( \sqrt{2x+2} + 2 \), we removed the square root and prepared the expression for further simplification.

Difference of Cubes

The difference of cubes is a special algebraic formula that allows you to factor expressions in the form of \( a^3 - b^3 \). The formula is:

• \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).
• This comes in handy for polynomial expressions when simplifying fractions or canceling terms.

In our exercise, the term \( x^3 - 1 \) was transformed using the difference of cubes formula.By expressing it as \( (x-1)(x^2 + x + 1) \), it made it easier to cancel out common factors, helping simplify the original equation.