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Chapter 3: Problem 68

Evaluate each limit by first converting each to a derivative at a particular \(x\) -value. $$\lim _{x \rightarrow-1} \frac{x^{2 / 9}-1}{x+1}$$

Short Answer

Expert verified
The limit evaluates to \(-\frac{2}{9}\).

Step by step solution

01

Recognize the Limit as a Derivative

The given limit resembles the definition of the derivative of a function at a point. Recall that the derivative of a function \( f(x) \) at a point \( a \) is given by \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \).
02

Identify the Function and the Point

Identify the function \( f(x) = x^{2/9} \) and recognize that we need \( f(-1) \) for the derivative. Substitute \( x = -1 \) into the function: \( f(-1) = (-1)^{2/9} = 1 \).
03

Apply Derivative Definition

Rewrite the original limit using the definition of the derivative for \( f(x) = x^{2/9} \) at \( x = -1 \): \[ \lim_{x \to -1} \frac{x^{2/9} - 1}{x + 1} = f'(-1). \]
04

Differentiate the Function

Differentiate \( f(x) = x^{2/9} \) using the power rule \( f'(x) = \frac{d}{dx}(x^{2/9}) = \frac{2}{9}x^{-7/9} \).
05

Evaluate the Derivative at the Particular X-Value

Substitute \( x = -1 \) into the derivative found in Step 4: \( f'(-1) = \frac{2}{9}(-1)^{-7/9} = \frac{2}{9} \). Since \((-1)^{-7/9} = -1\), the final result is \( f'(-1) = \frac{2}{9}( -1) = -\frac{2}{9} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit
The concept of a limit is foundational in calculus, acting as a bridge to understanding derivatives and integrals. Imagine you have a function, and you want to know what happens to its values as the input approaches a specific number. This is exactly what the limit seeks to explain.
  • Approaching a Value: When we say "the limit of a function as x approaches -1," we observe what happens to the function鈥檚 output as x gets closer and closer to -1, but not necessarily when x is exactly -1.
  • Simplifying Calculation: Limits help simplify situations where direct substitution of a value into a function might initially seem problematic, such as division by zero.
Understanding limits sets the stage for exploring how functions change, which is key for studying derivatives.
Derivative Definition
Understanding the definition of a derivative is crucial when working with calculus problems. The derivative fundamentally represents how a function changes at any given point, or more formally, the rate of change.
  • Basic Formula: The derivative of a function \( f(x) \) at a certain point \( a \) is given by the formula: \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
  • Use of Limit: As you can see, the derivative itself is defined using a limit. This means we're looking at the average rate of change of the function around a point, as the interval between values approaches zero.
Recognizing a limit expression as a derivative involves ensuring it resembles this fundamental definition.
Power Rule
One of the most powerful tools in differentiating functions is the Power Rule. Specifically, it provides a quick and easy method to find the derivative of a function where the variable is raised to a constant power.
  • Power Rule Explained: If you have a function like \( f(x) = x^n \), the derivative is found using the formula:\[ f'(x) = nx^{n-1} \]
  • Applying the Rule: In our exercise, the function \( f(x) = x^{2/9} \) means applying the Power Rule results in:\[ f'(x) = \frac{2}{9}x^{-7/9} \]
The Power Rule makes calculations straightforward and reduces complexities often encountered when finding derivatives.
Function Evaluation
Evaluating a function refers to finding the function's output value for a given input. This might seem straightforward, but it plays a vital role in derivative calculations.
  • Substitute Values: In our exercise, we substitute \( x = -1 \) into both the function and its derivative for direct evaluation.
  • Result of Evaluation: For \( f(x) = x^{2/9} \), substituting gives us \( f(-1) = 1 \), and evaluating its derivative yields \( f'(-1) = -\frac{2}{9} \).
This evaluation step confirms your derivative鈥檚 accuracy and is a key part of checking your work in calculus.

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Most popular questions from this chapter

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