Question

Evaluate: \(\operatorname{Lt}_{\mathrm{d} \rightarrow 3} \frac{\mathrm{d}^{3}-27}{\mathrm{~d}-3}\) (1) 3 (2) 9 (3) 27 (4) None of these

Short answer

To evaluate the limit \(\lim\limits_{d\to 3}\frac{d^3-27}{d-3}\), we first factor the numerator as a difference of cubes, resulting in \((d - 3)(d^2 + 3d + 3^2)\). This allows us to simplify the expression by canceling out the common term (d - 3) in the numerator and denominator, leaving us with the expression \((d^2 + 3d + 3^2)\). Then, we directly substitute the value of d to find the limit, resulting in the value 27. Thus, the correct answer is (3) 27.

Step-by-step solution

Factor the numerator

To factor the numerator, recognize that it's a difference of two cubes. The general formula for factoring a difference of cubes is given by: (a^3 - b^3) = (a - b)(a^2 + ab + b^2) In our case, a = d and b = 3, so the numerator can be factored into: \((d^3 - 3^3) = (d - 3)(d^2 + 3d + 3^2)\)

Simplify the expression

Now that we have factored the numerator, we can simplify the expression by cancelling out the common term from the numerator and denominator: \(\frac{d^3 - 3^3}{d - 3} = \frac{(d - 3)(d^2 + 3d + 3^2)}{d - 3}\) Cancel out the common term (d - 3) from the numerator and denominator to get: \((d^2 + 3d + 3^2)\)

Evaluate the limit

Since the expression has been simplified, we can now directly substitute the value of d to find the limit: \(\lim\limits_{d\to 3}(d^2 + 3d + 3^2) = (3^2 + 3\cdot 3 + 3^2)\) Calculate the expression to find the limit value: \((9 + 9 + 9) = 27\)

Choose the correct option

Now that we have evaluated the limit, we can compare our result with the given options: (1) 3 (2) 9 (3) 27 (4) None of these Our result matches option (3). So, the correct answer is (3) 27.

Key concepts

Difference of Cubes

In algebra, the difference of cubes refers to the expression of the form \(a^3 - b^3\). This is a special case of factoring which can simplify complex expressions, particularly beneficial when evaluating limits.
The formula for factoring the difference of cubes is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Here, you easily split your original cubic expression into a product of a linear and a quadratic expression.
For example, in our exercise, the difference of cubes is \(d^3 - 27\). By recognizing \(27\) as \(3^3\), we apply the formula:

• Let \(a = d\) and \(b = 3\).
• Substitute into the formula to factorize:

\((d - 3)(d^2 + 3d + 9)\).
Factoring helps in canceling terms in division problems, making calculus problems easier to manage.

Factorization

Factorization in mathematics is the process of breaking down a complex expression into simpler components that, when multiplied together, give the original expression.
Understanding how to factor expressions is crucial when solving calculus problems, especially when dealing with limits. In our problem, factorization was essential to simplify the expression \(\frac{d^3 - 27}{d - 3}\).
When we factor \(d^3 - 27\) using the difference of cubes formula, it allowed us to simplify the complex numerator into something more manageable.
Once factored, the expression was:

• \((d - 3)(d^2 + 3d + 9)\)

By factoring, the common term \((d - 3)\) could be canceled from the numerator and denominator, leading to a much simpler expression \(d^2 + 3d + 9\).
This shows how factorization aids in the simplification process, paving the way for easier limit evaluation.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus. When finding the limit of a function \(f(x)\) as \(x\) approaches a certain value, we are essentially determining what value the function converges to at that point.
In our exercise, we evaluate the limit \(\lim\limits_{d\to 3}\frac{d^3 - 27}{d - 3}\). After factorizing and simplifying the expression, we are left with \(d^2 + 3d + 9\).
Now, we simply substitute \(d = 3\) into the simplified expression and calculate:

• \(= 3^2 + 3\cdot 3 + 9\)
• \(= 9 + 9 + 9\)
• \(= 27\)

This straightforward numerical substitution technique makes it much simpler to evaluate limits after the expression has been properly simplified.

Calculus Problem Solving

Calculus problem solving involves applying various mathematical techniques to simplify and solve problems related to rates of change and accumulation. In the context of this exercise, solving the limit problem required both algebraic manipulation and calculus concepts.
The first step was recognizing and applying the difference of cubes formula. This allowed us to factor the numerator. Such problem-solving skills are essential for anyone learning calculus.
Next, the focus was on simplification through cancelling terms. This step is crucial, as complex fractions often obscure the limit evaluation.
Last was substituting the correct value to cleanly find the limit.

• Identify the correct formula or technique (e.g., difference of cubes).
• Simplify expressions wherever possible.
• Substitute to evaluate limits cleanly.

By following these logical steps, solving calculus problems becomes more manageable, helping students build a strong foundation in mathematical problem-solving.