Problem 36 - AP Calculus with 8 Practice Tests 15th

Question

\(\lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h}\) is (A) 0 (B) \(\frac{1}{12}\) (C) 1 (D) 192

Step-by-Step Solution

Answer

The answer is (B) \( \frac{1}{12} \).

Step 1: Identify the Expression

We have to find the limit as \( h \) approaches zero for the expression \( \lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h} \). This is equivalent to determining the derivative of \( f(x) = x^{1/3} \) at \( x = 8 \).

Step 2: Use Derivative Definition

The limit \( \lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h} \) is essentially the definition of a derivative. It computes the derivative of \( f(x) = x^{1/3} \) evaluated at \( x = 8 \).

Step 3: Apply the Chain Rule

The derivative of \( f(x) = x^{1/3} \) is found using the power rule \( f'(x) = \frac{1}{3}x^{-2/3} \). Substitute \( x = 8 \) to obtain the derivative at this point: \( f'(8) = \frac{1}{3}(8)^{-2/3} \).

Step 4: Simplify the Derivative

Calculate \( (8)^{-2/3} \):The cube root of 8 is 2, so \( 8^{1/3} = 2 \). This means \( 8^{-2/3} = \frac{1}{(8^{1/3})^2} = \frac{1}{2^2} = \frac{1}{4} \).

Step 5: Calculate Result

Now, substitute back into the expression for the derivative: \[ f'(8) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \]

Step 6: Conclude the Limit

Thus, the limit \( \lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h} \) is \( \frac{1}{12} \).

Key Concepts

Derivative Definition

In calculus, the derivative measures how a function changes as its input changes. It's like understanding the speed of a moving object. For any function \( f(x) \), the derivative \( f'(x) \) is defined as the limit:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

This represents how \( f(x) \) changes as \( x \) shifts slightly by \( h \). In essence, it's finding the slope of the tangent line at point \( x \) on the function.
In our exercise, the objective was to determine this limit specifically for the function \( x^{1/3} \), which represents the cube root. By setting \( x = 8 \) and applying the limit, we essentially find the derivative of the cube root function at that point.

Cube Roots

Cube roots involve finding a number which, when multiplied by itself three times, results in the given number. For example, the cube root of 8 is 2, because \( 2 \times 2 \times 2 = 8 \).
In our problem, we're dealing with the cube root expression \( \sqrt[3]{8+h} \). The goal is to understand how this expression behaves as \( h \) approaches zero, which is crucial for calculating the derivative.

• Cube roots can be expressed with the power notation: \( x^{1/3} \).
• This conversion helps simplify the process of differentiation, especially using the power rule.

Breaking complex cube root expressions into simpler polynomial forms can make limit and derivative calculations much more straightforward.

Power Rule

The power rule is a basic technique used to find the derivative of functions of the form \( x^n \). It's expressed simply as:

• If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

This rule saves time as it gives a direct formula for differentiation without needing the limit definition every time.
In our example, \( f(x) = x^{1/3} \), which can be differentiated using the power rule:

• The derivative: \( f'(x) = \frac{1}{3}x^{-2/3} \).

This makes it easy to calculate the rate of change at any point, and specifically at \( x = 8 \) in this problem. The power rule ensures we have a quick way to determine the slope of polynomial functions with fractional powers as well.

Chain Rule

The chain rule is a method in calculus for differentiating compositions of functions. When one function is inside another, the chain rule helps to take the derivative effectively. It's often written as:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

By multiplying the derivative of the outer function by that of the inner function, we can solve complex problems.
In our exercise, the function \( f(x) = x^{1/3} \) doesn't directly need the chain rule as it's a simple power, but understanding the chain rule is crucial for more complex integrations where nested functions appear.

• It's particularly useful when dealing with functions within functions, ensuring that every piece is correctly differentiated.

The power and chain rules together form a comprehensive toolkit for managing complex derivatives efficiently.