Problem 42 - Calculus Early Transcendentals

Question

(a) Use numerical and graphical evidence to guess the value of the limit $$\lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{x}-1}$$ (b) How close to 1 does \(x\) have to be to ensure that the function in part (a) is within a distance 0.5 of its limit?

Step-by-Step Solution

Answer

(a) The limit is 6. (b) \( x \) must be within 0.1 of 1.

Step 1: Simplify the Expression

First, observe the expression \( \frac{x^3 - 1}{\sqrt{x} - 1} \). Notice at \( x = 1 \), both the numerator and denominator become 0, causing an indeterminate form. We'll simplify by factoring the numerator. \( x^3 - 1 \) can be factored as \((x-1)(x^2 + x + 1)\). Now the expression is \( \frac{(x-1)(x^2 + x + 1)}{\sqrt{x} - 1} \).

Step 2: Rationalize the Denominator

To remove the indeterminate form, multiply the numerator and denominator by the conjugate of the denominator: \( \sqrt{x} + 1 \). This gives: \( \frac{(x-1)(x^2 + x + 1)(\sqrt{x} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \frac{(x-1)(x^2 + x + 1)(\sqrt{x} + 1)}{x - 1} \).

Step 3: Cancel Common Factors

Now, \( (x - 1) \) can be canceled from the numerator and denominator, simplifying to \( (x^2 + x + 1)(\sqrt{x} + 1) \). Evaluate the limit \( \lim_{x \to 1} ((x^2 + x + 1)(\sqrt{x} + 1)) \).

Step 4: Substitute and Calculate Limit

Now substitute \( x = 1 \) into the simplified expression: \((1^2 + 1 + 1)(\sqrt{1} + 1) = (3)(2) = 6\). Therefore, \( \lim_{x \to 1} \frac{x^3 - 1}{\sqrt{x} - 1} = 6 \).

Step 5: Define the Distance Requirement

To find how close \( x \) needs to be to 1 so that the expression is within a distance 0.5 of the limit, realize: \( \left| \frac{x^3 - 1}{\sqrt{x} - 1} - 6 \right| < 0.5 \).

Step 6: Express the Distance Inequality

Using the simplified expression from earlier, \( \left| (x^2 + x + 1)(\sqrt{x} + 1) - 6 \right| < 0.5 \). Solve this inequality to find the acceptable range for \( x \).

Step 7: Solve the Inequality

Since we know the limit is continuous around \( x = 1 \), we can approximate a small range around 1. After iterating through values or using numerical methods, observe that \( x \) must be within approximately 0.1 of 1 (i.e., \( 0.9 < x < 1.1 \)).

Key Concepts

Indeterminate forms

In calculus, indeterminate forms occur when an algebraic expression does not clearly resolve to a single number during evaluation. This often happens with limits. For example, if both the numerator and denominator of a fraction equal zero at the same point, we face an indeterminate form, such as \( \frac{0}{0} \). This form doesn't provide enough information about the actual value of the limit. We encountered this expression type in the exercise with \( \lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{x}-1} \), where both parts evaluated to zero at \( x = 1 \). Recognizing and correctly resolving indeterminate forms is vital since they can hide meaningful limit values.

Rationalizing

Rationalizing is a process used to eliminate irrational numbers from the denominator or to simplify expressions. In the given exercise, rationalizing helped transform the indeterminate form into a solvable expression. By multiplying both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{x}+1 \), we simplified the expression and made the zero terms explicit. This process turned \( \frac{x^3 - 1}{\sqrt{x} - 1} \) into \( \frac{(x-1)(x^2 + x + 1)(\sqrt{x} + 1)}{x - 1} \), allowing us to cancel \( (x - 1) \) and resolve the fraction. Rationalizing is a classic and handy technique in calculus for dealing with roots and achieving a clearer understanding of limits.

Numerical methods

Numerical methods involve approximating solutions to mathematical problems rather than finding exact answers through analytical approaches. When evaluating the limit \( \lim_{x \to 1} ((x^2 + x + 1)(\sqrt{x} + 1)) \), we can use numerical methods such as plugging values close to 1 into the function and observing the behavior. Sometimes, computations or plotting graphs with a calculator can reveal the behavior of a function near the limit point. In this case, observing values like 0.9, 0.95, 1.05, and 1.1 helps confirm that as \( x \to 1 \), the function approaches 6. Numerical methods are especially useful when algebraic manipulation is complex or doesn’t provide sufficient insight.

Continuity at a point

Continuity at a point essentially means that a function approaches and reaches a certain value without jumps or breaks. A function is continuous at a point \( x = c \) if three conditions are met: the function is defined at \( x = c \), the limit as \( x \) approaches \( c \) exists, and the limit equals the function's value at that point. In our exercise, once simplified, the expression \( (x^2 + x + 1)(\sqrt{x} + 1) \) was shown to be a continuous function at \( x = 1 \), thereby confirming our calculated limit of 6 for \( \lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{x}-1} \). Conceptually, understanding continuity helps anticipate points within a function where behavior might change or remain steady.