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Magazine Chapter 3: Problem 36
\(\lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h}\) is (A) 0 (B) \(\frac{1}{12}\) (C) 1 (D) 192
Short Answer
Expert verified
The answer is (B) \( \frac{1}{12} \).
Step by step solution
01
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 \( x = 8 \).
02
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 \( x = 8 \).
03
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: \( f'(8) = \frac{1}{3}(8)^{-2/3} \).
04
Simplify the Derivative
Calculate \( (8)^{-2/3} \):The cube root of 8 is 2, so \( 8^{1/3} = 2 \). This means \( 8^{-2/3} = \frac{1}{(8^{1/3})^2} = \frac{1}{2^2} = \frac{1}{4} \).
05
Calculate Result
Now, substitute back into the expression for the derivative: \[ f'(8) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \]
06
Conclude the Limit
Thus, the limit \( \lim _{h \rightarrow 0} \frac{\sqrt[3]{8+h}-2}{h} \) is \( \frac{1}{12} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
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.
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.
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:
In our example, \( f(x) = x^{1/3} \), which can be differentiated using the power rule:
- If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
In our example, \( f(x) = x^{1/3} \), which can be differentiated using the power rule:
- The derivative: \( f'(x) = \frac{1}{3}x^{-2/3} \).
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:
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.
- If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
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.
