Problem 35 - AP Calculus with 8 Practice Tests 15th

Question

\(\lim _{h \rightarrow 0} \frac{(1+h)^{6}-1}{h}\) is (A) 0 (B) 1 (C) 6 (D) nonexistent

Step-by-Step Solution

Answer

The limit is 6.

Step 1: Understand the Situation

We are asked to find the limit of the expression \( \lim_{h \to 0} \frac{(1+h)^6 - 1}{h} \). This expression involves the first derivative of the function \( f(x) = x^6 \) at \( x = 1 \), based on the definition of the derivative.

Step 2: Expand Using Binomial Theorem

Use the binomial theorem to expand \((1+h)^6\). According to the binomial theorem: \((1+h)^6 = 1 + 6h + 15h^2 + 20h^3 + 15h^4 + 6h^5 + h^6\).

Step 3: Simplify the Expression

Substitute the expansion into the limit expression: \(\frac{(1+h)^6 - 1}{h} = \frac{[1 + 6h + 15h^2 + 20h^3 + 15h^4 + 6h^5 + h^6] - 1}{h}\). This results in: \(\frac{6h + 15h^2 + 20h^3 + 15h^4 + 6h^5 + h^6}{h}\).

Step 4: Cancel out \( h \)

Divide each term by \(h\): \(6 + 15h + 20h^2 + 15h^3 + 6h^4 + h^5\).

Step 5: Evaluate the Limit

Now, take the limit as \(h\) approaches 0: \(\lim_{h \to 0} (6 + 15h + 20h^2 + 15h^3 + 6h^4 + h^5)\). All terms with \(h\) will vanish, leaving us with \(6\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in algebra that allows us to expand expressions raised to a power. For example, when we expand \((1+h)^n\), the theorem provides a formula to calculate each term in the expansion:

• The first term is always 1.
• The second term is the power \(n\) times \(h\), which reflects how many different ways we can arrange \(h\).
• Subsequent terms involve binomial coefficients, which can be derived from Pascal's triangle or calculated using combinations formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

In our exercise, we applied the binomial theorem to \((1+h)^6\), resulting in the expansion: \[1 + 6h + 15h^2 + 20h^3 + 15h^4 + 6h^5 + h^6\]. This expansion simplifies the expression by breaking it down into easier components to work with.

Derivative

The concept of a derivative is fundamental in calculus. It captures how a function changes as its input changes, specifically the function's rate of change.
In essence, the derivative of a function at a certain point informs us about the slope of the tangent line to the function at that point.
To find the derivative of a function, we often use the limit definition of a derivative. This involves taking the limit of the average rate of change of the function as the interval approaches zero.
For a function \(f(x) = x^6\), the derivative, \(f'(x)\), at any point \(x\) reflects how steeply \(x^6\) is changing.
For instance, at \(x = 1\), the definition is found using the limit \[ \lim_{h \to 0} \frac{(1+h)^6 - 1}{h}\]. This gives us insight into the function's behavior near \(x = 1\).

Limit Evaluation

The process of limit evaluation is a crucial aspect of calculus, providing insights into the behavior of functions as inputs approach certain values. When evaluating limits, particularly those involving expressions like \(\frac{f(x)}{g(x)}\) as \(x\) approaches a specific value, we seek to understand the function's behavior near that point.

• If the evaluation directly yields an indeterminate form like \(\frac{0}{0}\), further simplification is necessary.
• Some common techniques involve factorization, rationalization, and applying theorems such as l'Hôpital's rule.

In our exercise, the initial fraction was simplified using an expansion obtained from the binomial theorem.
This simplification eliminated the potential for an indeterminate form and made the limit approachable.

Limit Simplification

Simplifying limits often involves breaking down expressions into simpler components, making them manageable and easier to evaluate. In the case of limits resulting in \(\frac{0}{0}\) indeterminate forms, simplification is crucial.
Generally, this means rewriting expressions so direct substitution becomes possible. In our example, the expression \(\frac{(1+h)^6 - 1}{h}\) led us to expand using the binomial theorem, allowing each term involving \(h\) to be divided by \(h\).
As a result, cancellation of \(h\) occurred, thus simplifying the limit expression to a form where direct evaluation was possible.
Finally, after removing all terms with \(h\) from the simplified form, the limit was easily found: \[ \lim_{h \to 0} (6 + 15h + 20h^2 + 15h^3 + 6h^4 + h^5) = 6 \], as all terms except the constant 6 vanished when \(h\) approached zero, concluding the simplification process.