Chapter 1: Problem 48
A positive number \(\epsilon\) and the limit \(L\) of a function \(f\) at \(-\infty\)
are given. Find a negative number \(N\) such that \(|f(x)-L|<\epsilon\) if \(x
Short Answer
Expert verified
Choose \( N = -1001 \).
Step by step solution
01
Identify the Limit Function
The given limit is \( \lim_{x \rightarrow -\infty} \frac{x}{x+1} = 1 \). We need to use this expression to find \( N \) such that for all \( x < N \), the condition \( |f(x) - L| < \epsilon \) is satisfied, where \( L = 1 \) and \( \epsilon = 0.001 \).
02
Express the Error Inequality
We want \( \left| \frac{x}{x+1} - 1 \right| < 0.001 \). This inequality will help us to determine how \( x \) must be related to \( N \).
03
Simplify the Expression
By simplifying the expression \( \left| \frac{x}{x+1} - 1 \right| \), we have: \( \frac{x}{x+1} - 1 = \frac{x - (x+1)}{x+1} = \frac{-1}{x+1} \). Therefore, the inequality becomes \( \left| \frac{-1}{x+1} \right| < 0.001 \).
04
Set Up the Absolute Inequality
The expression \( \left| \frac{-1}{x+1} \right| < 0.001 \) simplifies to \( \frac{1}{|x+1|} < 0.001 \), which further simplifies to \( |x+1| > 1000 \) when multiplied by 1000.
05
Solve the Inequality for x
For \( |x+1| > 1000 \), we have two cases: 1. \( x+1 > 1000 \), leading to \( x > 999 \) and 2. \( x+1 < -1000 \), leading to \( x < -1001 \). Since we need \( x < N \), we select the second case, which gives \( x < -1001 \). Therefore, we can choose \( N = -1001 \).
06
Verification
Check if \( N = -1001 \) satisfies the condition. For any \( x < -1001 \), the absolute difference is \( \left| \frac{-1}{x+1} \right| < 0.001 \), confirming our selection of \( N \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Epsilon-Delta Definition of a Limit
When exploring the concept of limits, particularly limits at infinity, the epsilon-delta definition provides a rigorous framework. In simple terms, for a function \( f(x) \), the limit \( L \) as \( x \) approaches infinity or negative infinity is found by ensuring that \( f(x) \) gets arbitrarily close to \( L \). Here, "arbitrarily close" is defined by a small positive number \( \epsilon \), which represents how close we want our function value to get to the limit.
The formal definition states: For every \( \epsilon > 0 \), there exists a \( N \) such that if \( x < N \) (for limits at negative infinity) or \( x > N \) (for limits at positive infinity), then \( |f(x) - L| < \epsilon \).
This concept essentially boils down to:
The formal definition states: For every \( \epsilon > 0 \), there exists a \( N \) such that if \( x < N \) (for limits at negative infinity) or \( x > N \) (for limits at positive infinity), then \( |f(x) - L| < \epsilon \).
This concept essentially boils down to:
- Choosing any small tolerance value, \( \epsilon \).
- Determining a point, \( N \), beyond which the function value remains within \( \epsilon \) of \( L \).
Inequalities in Calculus
Inequalities are critical tools in calculus, especially when working with limits and the epsilon-delta definition. They are used to describe relationships between numbers and, in this context, help us determine the behavior of functions concerning their limits.
To solve the inequality \( \left| \frac{-1}{x+1} \right| < 0.001 \), we can interpret it in the following steps:
To solve the inequality \( \left| \frac{-1}{x+1} \right| < 0.001 \), we can interpret it in the following steps:
- The expression inside the absolute value becomes an inequality: \( \frac{-1}{x+1} < 0.001 \) and \( \frac{-1}{x+1} > -0.001 \).
- Simplify the inequality by eliminating the absolute value and solving the inequality for \( x \).
- From these computations, determine \( N \) such that when \( x < N \), the inequality is satisfied.
Absolute Value
The concept of absolute value is important in calculus, providing a way to describe distance on the real number line, regardless of direction. When we see absolute value signs, \(|...|\), it signifies how far a number is from zero.
Using absolute value in inequalities helps measure how close one expression is to another, such as \(|f(x)-L|\). In the problem \( |\frac{x}{x+1} - 1| < 0.001 \), we wish to keep the expression \( \frac{x}{x+1} \) close to its limit \( L = 1 \).
When simplifying and solving:
Using absolute value in inequalities helps measure how close one expression is to another, such as \(|f(x)-L|\). In the problem \( |\frac{x}{x+1} - 1| < 0.001 \), we wish to keep the expression \( \frac{x}{x+1} \) close to its limit \( L = 1 \).
When simplifying and solving:
- Start by recognizing absolute value as a distance measure.
- Interpret \( \left| x-a \right| < b \) as a range: \( -b < x-a < b \).
- Use this understanding to set up inequalities that establish the permissible range of \( x \).