Problem 31 - Calculus

Question

Use algebra to evaluate the limit. $$\lim _{h \rightarrow 0} \frac{(1+h)^{3}-1}{h}$$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: Expand the Expression

Start by expanding the numerator \( (1+h)^3 - 1 \). Use the binomial theorem or the formula for cubes: \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). Substituting \( a = 1 \) and \( b = h \), we have: \( (1+h)^3 = 1 + 3h + 3h^2 + h^3 \). Therefore, the expression becomes \( (1 + 3h + 3h^2 + h^3) - 1 = 3h + 3h^2 + h^3 \).

Step 2: Simplify the Fraction

Substitute the expanded expression back into the limit: \(\lim_{h \to 0} \frac{3h + 3h^2 + h^3}{h} \). Simplify the fraction by factoring \h\ out from the numerator: \( 3h + 3h^2 + h^3 = h(3 + 3h + h^2) \). The fraction simplifies to \( \frac{h(3 + 3h + h^2)}{h} = 3 + 3h + h^2 \).

Step 3: Evaluate the Limit

Now that the expression is simplified, evaluate the limit as \( h \) approaches 0. Substitute \( h = 0 \) into \( 3 + 3h + h^2 \), which results in \( 3 + 3\cdot0 + 0^2 = 3 \). Therefore, the limit is \( 3 \).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool used to expand expressions raised to a power. In the context of this problem, we use it to expand \((1+h)^3\). According to the binomial theorem, any binomial expression can be expanded as a series involving combinations and powers of its terms. Specifically, when expanding \((a+b)^n\), the form is given by:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

For the problem at hand, we let \(a = 1\) and \(b = h\). Using the expansion for a cube:

• \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

So, when applying this to our expression, it becomes easy to manage and allows us to move on to the next steps of the problem.

Expression Expansion

Expression expansion is taking a compact mathematical expression and rewriting it in a longer form. This is useful for simplifying problems, especially when dealing with limits and derivatives.
In the case of our limit problem, once we use the binomial theorem to expand \((1+h)^3\), we get:

• \((1 + 3h + 3h^2 + h^3)\)

From here, we can easily subtract \(1\), resulting in:

• \((3h + 3h^2 + h^3)\)

This is a crucial step because it transforms a potentially complicated fraction into something we can simplify and easily solve for a limit as \(h\) approaches zero.

Limit Simplification

Simplifying an expression before placing it into a limit can make solving the limit much clearer. After expanding the numerator, we simplify the fraction:\(\lim_{h \rightarrow 0} \frac{3h + 3h^2 + h^3}{h}\)By factoring out \(h\) from the numerator, the expression simplifies:

• \(3h + 3h^2 + h^3 = h(3 + 3h + h^2)\)

Then, the fraction reduces to:

• \(\frac{h(3 + 3h + h^2)}{h} = 3 + 3h + h^2\)

This simplification eliminates the fraction, allowing us to directly substitute \(h = 0\) in the resulting polynomial, making the limit straightforward to evaluate.

Algebraic Manipulation

Algebraic manipulation refers to using algebraic rules and operations to simplify or rearrange expressions. In this exercise, careful manipulation enables us to simplify the fraction and easily evaluate the limit.
By factoring \(h\) from the terms in the numerator, we address the potential for undefined expressions when \(h\) approaches zero. Canceling \(h\) from the numerator and denominator is fundamental here.
Once we have simplified to the expression \(3 + 3h + h^2\), evaluating the limit becomes an easy task.

• We substitute \(h = 0\)
• This results in \(3+0+0 = 3\)

Thus, understanding how to manipulate expressions is vital for smoothly assessing limits and handling potentially undefined behavior.