Problem 55 - Calculus: Early Transcendentals

Question

(a) Use numerical and graphical evidence to guess the value of the limit $$ \lim_{x \to 1}\frac{x^3 - 1}{\sqrt{x} - 1} $$ (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

The limit is 6. Ensure \( x \) is between roughly 0.9 and 1.1.

Step 1: Simplify the Expression

We begin by simplifying the expression \( \frac{x^3 - 1}{\sqrt{x} - 1} \). Notice that the expression can lead to the indeterminate form \( \frac{0}{0} \) when \( x \) approaches 1. To handle this, we factor the numerator using the difference of cubes: \( x^3 - 1 = (x-1)(x^2 + x + 1) \). Thus, the expression becomes \( \frac{(x-1)(x^2 + x + 1)}{\sqrt{x} - 1} \).

Step 2: Rationalize the Denominator

To eliminate the square root in the denominator, we multiply and divide by the conjugate: \( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \). Multiply the entire expression by \( \frac{\sqrt{x} + 1}{\sqrt{x} + 1} \), giving us \( \frac{(x-1)(x^2 + x + 1)(\sqrt{x} + 1)}{x - 1} \). This simplifies to \( (x^2 + x + 1)(\sqrt{x} + 1) \).

Step 3: Substitute and Evaluate the Limit

Now, substitute \( x = 1 \) directly into the simplified expression: \( (1^2 + 1 + 1)(\sqrt{1} + 1) = (3)(2) = 6 \). This value indicates the limit. The expression is now valid for substitution since it no longer has an indeterminate form.

Step 4: Determine the Distance Condition

The task is to find an interval where the function is within 0.5 units from its limit, 6. We need \( |f(x) - 6| < 0.5 \). Simplify this to find \(-0.5 < f(x) - 6 < 0.5\), or equivalently, \(5.5 < f(x) < 6.5\). Find the \( x \) values that satisfy this condition.

Step 5: Solve the Inequality

Using the simplified function \( (x^2 + x + 1)(\sqrt{x} + 1) \), solve the inequality \(5.5 < (x^2 + x + 1)(\sqrt{x} + 1) < 6.5\). Approximating numerically around \( x = 1 \) and calculating values for \( x = 0.9 \) and \( x = 1.1 \), confirms it stays within the interval. Thus, ensure \( x\) is around 0.9 to 1.1, by numerical testing, for satisfying the limit distance condition.

Key Concepts

Numerical Approximation

When dealing with calculus limit problems, numerical approximation becomes a handy tool. It involves calculating and estimating the value of a limit numerically. This can be done by inputting values close to the limit point into the function and observing the outputs.

• For example, if approaching 1, we might evaluate the function at values like 0.9, 0.95, 1.05, and 1.1.
• The idea is to check whether outputs are converging to a single number.

This approach can give students a practical understanding of limits, by showing how the function behaves near the point we're interested in.
It is important to choose values close to the target to see the true behavior closely.

Graphical Analysis

Graphical analysis offers a visual method to understand and predict the behavior of functions. By plotting the function in question, we can see how it behaves as it approaches a particular point. Graphs can often reveal asymptotic behavior, discontinuities, or convergence that might not be obvious algebraically.

• Graph as many points as possible near the limit to see the trend.
• It's common to use graphing calculators or software to achieve a precise representation.

Visual representation can strongly reinforce the numerical approximation by confirming or questioning the numerical findings. Many times, the graph makes the concept of limits more intuitive.

Difference of Cubes

The difference of cubes is a special factoring formula used to simplify expressions of the form \(a^3 - b^3\).
In our problem, it was crucial for simplifying the numerator \(x^3 - 1\). The formula is:\[a^3 - b^3 = (a-b)(a^2 + ab + b^2)\]For our exercise, \(x^3 - 1\) can be factored as:\[x^3 - 1 = (x-1)(x^2 + x + 1)\]This factoring removed the indeterminate form issue, allowing further simplification and calculation of the limit. Understanding such algebraic identities is crucial in simplifying complex limit expressions.

Indeterminate Forms

Indeterminate forms are expressions that don't provide enough information to directly determine the limit. The most common forms encountered in calculus are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).
In this problem, the original expression \(\frac{x^3 - 1}{\sqrt{x} - 1}\) creates a \(\frac{0}{0}\) situation as \(x\) approaches 1. Such forms require special handling, either through algebraic manipulation, such as factoring or rationalizing, or L'Hôpital's rule for differentiation.
Resolving an indeterminate form is a crucial step to successfully evaluate a limit. This often involves creative algebraic strategies that simplify the expression into a determinate form, allowing for normal evaluation.