Problem 3 - Brief Applied Calculus

Question

Complete the tables and use them to find the given limits. Round calculations to three decimal places. A graphing calculator with a TABLE feature will be very helpful. a. \(\lim _{x \rightarrow 1^{-}}\left(\frac{x^{3}-1}{x-1}\right)\) b. \(\lim _{x \rightarrow 1^{+}}\left(\frac{x^{3}-1}{x-1}\right)\) c. \(\lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)\)

Step-by-Step Solution

Answer

All limits are equal to 3.

Step 1: Analyze the Function

The function we are dealing with is \( f(x) = \frac{x^3 - 1}{x - 1} \). This is a rational function which is undefined at \( x = 1 \) because the denominator becomes zero.

Step 2: Simplify the Function

Notice that the numerator can be factored using the difference of cubes formula: \( x^3 - 1 = (x-1)(x^2 + x + 1) \). This allows the function to be simplified as long as \( x eq 1 \):\[\frac{x^3 - 1}{x - 1} = x^2 + x + 1\] when \( x eq 1 \).

Step 3: Evaluate the Limit from the Left (x -> 1^−)

To find \( \lim_{x \to 1^-} \), we analyze values very close to 1 from the left. Using the simplified function \( x^2 + x + 1 \), substitute values such as \( x = 0.9, 0.99, 0.999 \) to approximate the limit:- \( f(0.9) = 0.9^2 + 0.9 + 1 = 2.71 \)- \( f(0.99) = 0.99^2 + 0.99 + 1 = 2.9701 \)- \( f(0.999) = 0.999^2 + 0.999 + 1 = 2.997001 \)The results suggest the limit from the left is approaching 3.

Step 4: Evaluate the Limit from the Right (x -> 1^+)

For \( \lim_{x \to 1^+} \), we analyze values very close to 1 from the right using the simplified function \( x^2 + x + 1 \):- \( f(1.1) = 1.1^2 + 1.1 + 1 = 3.31 \)- \( f(1.01) = 1.01^2 + 1.01 + 1 = 3.0301 \)- \( f(1.001) = 1.001^2 + 1.001 + 1 = 3.003001 \)The results suggest the limit from the right is also approaching 3.

Step 5: Evaluate the Overall Limit (x -> 1)

Since both the left-hand limit \( \lim_{x \to 1^-} \) and the right-hand limit \( \lim_{x \to 1^+} \) are equal to 3, we conclude that the overall limit exists and is:\( \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = 3 \).

Key Concepts

Rational Functions

Rational functions are an essential part of calculus. They represent the ratio of two polynomials. In our example, the function is given by \[ f(x) = \frac{x^3 - 1}{x - 1} \]. This function is undefined when the denominator becomes zero, which happens at \( x = 1 \). This means that you cannot directly substitute \( x = 1 \) into the function. You would get a zero in the denominator, leading to an undefined expression.
To work around this, we look for ways to simplify the function. Often, rational functions can be simplified by factoring the numerator or denominator, cancelling out terms, and making the expression easier to manage, just like in this problem.
Understanding how rational functions behave is crucial because they often show up in real-world applications, like in physics and economics. If you know how to handle these functions, calculus problems become a lot more manageable.

Difference of Cubes

The difference of cubes formula is a handy tool in algebra. It allows us to factor expressions like \( x^3 - 1 \). The formula for the difference of cubes is \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]. In our exercise, we set \( a = x \) and \( b = 1 \), so we apply the formula to get \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \].
Using this factored form, we can cancel the \( x - 1 \) in the numerator and denominator, providing a simplified function: \[ \frac{x^3 - 1}{x - 1} = x^2 + x + 1 \], as long as \( x eq 1 \).
This simplification is what makes evaluating limits possible, as it helps remove the term that causes the function to be undefined. This technique is commonly used to solve limits, especially when direct substitution initially leads to an indeterminate form.

Evaluating Limits

Evaluating limits is a fundamental concept in calculus, used to understand the behavior of functions as they approach a specific point. In problems like ours, direct substitution doesn't work initially because it leads to an undefined expression. Hence, we use algebraic manipulation like factoring.
It's important to consider limits from both directions because they must equal the same value for the overall limit to exist at that point. In this problem, we looked at \[ \lim_{x \to 1^-} \] and \[ \lim_{x \to 1^+} \] to make sure that both approached the same number.

• From the left: As \( x \) gets closer to 1 from the left (like 0.999), we end up with values very close to 3.
• From the right: As \( x \) nears 1 from the right (such as 1.001), the values again approach 3.

With both limits matching, we confirm that the overall limit is 3: \[ \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = 3 \].
This technique of evaluating limits is foundational for more advanced calculus topics, such as derivatives and integrals.