Problem 20 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{h \rightarrow 0} \frac{(2+h)^{3}-8}{h}$$

Step-by-Step Solution

Answer

The limit is 12.

Step 1: Recognize as a Derivative

The expression \( \lim_{h \to 0} \frac{(2+h)^3 - 8}{h} \) resembles the definition of a derivative \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). Here, the function \( f(x) = x^3 \) and \( a = 2 \), so this limit represents the derivative of \( f \) at \( x = 2 \).

Step 2: Expand the Term \( (2 + h)^3 \)

Using the binomial theorem or straightforward expansion, expand \( (2+h)^3 = 8 + 12h + 6h^2 + h^3 \).

Step 3: Substitute the Expanded Expression

Substitute \((2+h)^3 = 8 + 12h + 6h^2 + h^3\) back into the limit: \[\lim _{h ightarrow 0} \frac{(8 + 12h + 6h^2 + h^3) - 8}{h}.\]

Step 4: Simplify the Limit Expression

Cancel out the 8s and combine terms:\[\lim _{h ightarrow 0} \frac{12h + 6h^2 + h^3}{h} = \lim _{h ightarrow 0} (12 + 6h + h^2).\]

Step 5: Evaluate the Limit

As \( h \to 0 \), the terms involving \( h \) tend to zero: \[\lim _{h ightarrow 0} (12 + 6h + h^2) = 12.\]

Key Concepts

Understanding Derivatives

In calculus, derivatives represent the rate at which a function changes at any given point. This concept is crucial, as it allows us to understand how a function behaves by looking at its slope at particular points. In simple terms, the derivative of a function at a point gives us the slope of the tangent line to the curve at that point.

To find a derivative, we often use the limit of a difference quotient, which is exactly what we see in our example: \[\lim _{h \rightarrow 0} \frac{(2+h)^{3}-8}{h}\]
This form resembles the definition of a derivative, \(f'(a)\), written as:\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.\]

• The expression involves \((2+h)^3\), showing that it’s related to the cubic function \(f(x) = x^3\).
• We are searching for the derivative at \(x = 2\), which can help us understand how quickly \(x^3\) changes at this point.

By recognizing the function and plugging into its derivative form, you gain insights into the changing nature of functions through calculus.

Exploring the Binomial Theorem

The binomial theorem is a powerful tool for expanding expressions of the form \((a+b)^n\). It allows us to break down polynomial expressions into simpler components that are easier to handle.

In our example, we expand \((2 + h)^3\) using this theorem or straightforward polynomial expansion. The binomial theorem provides a formula to achieve this:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]For our case, \((2+h)^3\), the expansion becomes:

• \(2^3 = 8\)
• \(3 \cdot 2^2 \cdot h = 12h\)
• \(3 \cdot 2 \cdot h^2 = 6h^2\)
• \(h^3\) as \(8 + 12h + 6h^2 + h^3\)

This breakdown helps us substitute back into the original limit expression, simplifying our calculations. This demonstrates why the binomial theorem is such a valuable tool in calculus and algebra.

Polynomial Expansion Simplified

Polynomial expansion simply refers to expressing a polynomial power in its full expanded form. In the example, we transform the term \((2+h)^3\) into a more manageable sum.

When expanded fully, \((2+h)^3\) results in:

• \(8\) representing \(2^3\)
• \(12h\) derived from multiplying coefficients
• \(6h^2\)
• \(h^3\)

Each term is the result of multiplying the different parts of the original expression, helping you match them to their respective powers and coefficients. This expanded form is crucial for plugging back into our expression to simplify the final result for derivatives.

Understanding polynomial expansion helps in tackling complex expressions, especially when evaluating limits or derivatives.

The Role of Calculus in Problem Solving

Calculus is the field of mathematics that deals with continuous change. Two of its most crucial operations are finding derivatives and integrals. In this example, solving the limit is rooted in derivative calculation, which is a fundamental aspect of calculus.

Calculus allows us to:

• Understand how functions behave by analyzing their growth rate.
• Manage complex problems by breaking them down using derivatives or integrals.
• Link real-world problems, like motion from physics, to mathematical functions through derivatives.

In this scenario, the limit \( \lim _{h \rightarrow 0} \frac{(2+h)^3 - 8}{h} \) exemplifies how calculus aids in deriving instantaneous rate of change. As \(h\) approaches zero, any terms involving \(h\) reduce to zero, simplifying the arithmetic to calculate the behavior of \((2+h)^3\) around \(x=2\).

Calculus is integral for understanding the intricate relationships within mathematics, providing the language and tools necessary for solving dynamic problems.