Problem 46 - Calculus

Question

Use algebra to evaluate the limits. $$\lim _{h \rightarrow 0} \frac{(2-h)^{3}-8}{h}$$

Step-by-Step Solution

Answer

The limit is -12.

Step 1: Expand the Binomial

To evaluate the limit, start by expanding the binomial expression \( (2-h)^3 \). Use the binomial theorem, where \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):\\((2-h)^3 = 2^3 - 3 imes 2^2 imes h + 3 imes 2 imes h^2 - h^3 = 8 - 12h + 6h^2 - h^3\).

Step 2: Substitute and Simplify the Expression

Substitute the expanded expression back into the limit: \[\lim_{h \rightarrow 0} \frac{(2-h)^3 - 8}{h} = \lim_{h \rightarrow 0} \frac{8 - 12h + 6h^2 - h^3 - 8}{h}\]Now simplify:\[\lim_{h \rightarrow 0} \frac{-12h + 6h^2 - h^3}{h}\].

Step 3: Factor Out the Common Term

Factor out \(h\) from the numerator: \[\lim_{h \rightarrow 0} \frac{-12h + 6h^2 - h^3}{h} = \lim_{h \rightarrow 0} (-12 + 6h - h^2)\].

Step 4: Evaluate the Limit as h Approaches 0

Now substitute \(h = 0\) into the expression:\[\lim_{h \rightarrow 0} (-12 + 6h - h^2) = -12 + 6(0) - (0)^2 = -12\].

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra for expanding expressions that are raised to a power, specifically binomials. A binomial is simply an expression that has two terms, like \((a + b)\). The theorem provides a formula to expand expressions like \((a + b)^n\), where \(n\) is a positive integer.
The general form of the Binomial Theorem is given by:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]Here, \(\binom{n}{k}\) stands for a binomial coefficient, which is calculated as \(\frac{n!}{k!(n-k)!}\). This formula allows us to expand the expression into a series of terms.
For example, expanding \((2 - h)^3\) using the Binomial Theorem involves using the formula:

• \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\)
• Where \(a = 2\) and \(b = h\).

By substituting the values, we get:\( (2 - h)^3 = 8 - 12h + 6h^2 - h^3 \).
This step simplifies calculations in problems like limit evaluation, making it easier to proceed with further steps of algebraic manipulation.

Limit Evaluation

Limit evaluation is an essential concept in calculus that deals with finding the value that a function approaches as the input approaches some value. In problems like \(\lim_{h \rightarrow 0} \frac{(2-h)^3 - 8}{h}\), the goal is to evaluate the function's behavior as \(h\) gets very close to 0.
Evaluating such limits often involves algebraic manipulation to simplify the expression such that direct substitution becomes feasible. After expanding and simplifying the expression, you substitute the variable to find the limit value.
For the expression\[\lim_{h \rightarrow 0} \frac{-12h + 6h^2 - h^3}{h}\]you can simplify by factoring out the \(h\) from the numerator:\[\lim_{h \rightarrow 0} (-12 + 6h - h^2)\]Once the factor is removed, you can replace \(h\) with 0, leading to the final solution:\(-12\).
This straightforward method, when executed properly, reveals the behavior of functions around particular points. Limits form the foundation for many concepts in calculus, like derivatives and continuity.

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial as a sum of terms. Each term is derived from multiplying different powers of the variables involved, often using specific algebraic rules or formulas. In calculus, polynomial expansion can simplify complex expressions and make them more manageable for further analysis, like limit evaluation.
For instance, expanding \((2 - h)^3\) involves distributing and simplifying each term according to the powers of \(h\). This leads to a polynomial form:\(2^3 - 3 \cdot 2^2 \cdot h + 3 \cdot 2 \cdot h^2 - h^3\).
Such expanded forms can usually be divided or arranged easily due to their simple structure, aiding in calculus operations such as finding limits or derivatives.

• Helps break down expressions into simpler components.
• Facilitates deeper analysis and manipulation in problems.

In the given exercise, polynomial expansion allowed for the algebraic simplification necessary to evaluate the limit. This demonstrates how crucial understanding this concept is in tackling calculus problems effectively.