Chapter 1: Problem 45
A positive number \(\epsilon\) and the limit \(L\) of a function \(f\) at \(+\infty\) are given. Find a positive number \(N\) such that \(|f(x)-L|<\epsilon\) if \(x>N\). $$ \lim _{x \rightarrow+\infty} \frac{x}{x+1}=1 ; \epsilon=0.001 $$
Short Answer
Expert verified
Choose \( N = 999 \).
Step by step solution
01
Identify the Function and Limit
The function given is \( f(x) = \frac{x}{x+1} \) and the limit as \( x \) approaches infinity is \( L = 1 \). We are tasked with ensuring the difference \( |f(x) - L| < \epsilon = 0.001 \) for all \( x > N \).
02
Express the Absolute Difference
Calculate the absolute difference between the function and its limit: \[|f(x) - L| = \left| \frac{x}{x+1} - 1 \right| = \left| \frac{x - (x+1)}{x+1} \right| = \left| \frac{-1}{x+1} \right| = \frac{1}{x+1}. \]
03
Set Up the Inequality
We need \( \frac{1}{x+1} < 0.001 \). This will ensure \( |f(x) - L| < \epsilon \).
04
Solve the Inequality for \( x \)
Solve the inequality \( \frac{1}{x+1} < 0.001 \):- Multiplying both sides by \( x+1 \) (assuming \( x+1 > 0 \)) gives: \[ 1 < 0.001(x+1) \]- Divide both sides by 0.001: \[ x+1 > 1000 \]- Subtract 1 from both sides: \[ x > 999 \].Thus, \( N = 999 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Epsilon-Delta Definition
The Epsilon-Delta Definition is a cornerstone of calculus used to define the concept of limits rigorously. It describes a precise way to discuss how a function behaves as the input approaches a certain value.
Consider a function \( f(x) \) and a limit \( L \). If for every positive number \( \epsilon \) there exists a corresponding positive number \( \delta \) such that:
For the limit at infinity, as in our exercise, \( \epsilon \) is given (0.001 in this case), and we need to find \( N \) such that \( |f(x) - L| < \epsilon \) for all \( x > N \). This is an adaptation of the epsilon-delta concept to handle limits as \( x \) heads towards infinity.
Consider a function \( f(x) \) and a limit \( L \). If for every positive number \( \epsilon \) there exists a corresponding positive number \( \delta \) such that:
- Whenever \( 0 < |x-a| < \delta \), it implies \( |f(x) - L| < \epsilon \)
For the limit at infinity, as in our exercise, \( \epsilon \) is given (0.001 in this case), and we need to find \( N \) such that \( |f(x) - L| < \epsilon \) for all \( x > N \). This is an adaptation of the epsilon-delta concept to handle limits as \( x \) heads towards infinity.
Continuous Functions
Continuous functions are those that do not have any abrupt changes in value—they are smooth without breaks or jumps. A function \( f(x) \) is continuous at a point \( a \) if:
The function \( f(x) = \frac{x}{x+1} \) from the exercise is continuous because the operations involved (division by \( x+1 \)) do not introduce discontinuities as \( x \) approaches infinity. This continuous behavior is crucial when analyzing the function near its limit at infinity.
Even as \( x \rightarrow +\infty \), continuity ensures the function moves smoothly toward the limit \( L = 1 \), making it easier to apply epsilon-delta rules and find an \( N \) such that \( |f(x) - L| < \epsilon \).
- \( f(a) \) is defined
- The limit \( \lim_{x \to a}f(x) \) exists
- \( \lim_{x \to a}f(x) = f(a) \)
The function \( f(x) = \frac{x}{x+1} \) from the exercise is continuous because the operations involved (division by \( x+1 \)) do not introduce discontinuities as \( x \) approaches infinity. This continuous behavior is crucial when analyzing the function near its limit at infinity.
Even as \( x \rightarrow +\infty \), continuity ensures the function moves smoothly toward the limit \( L = 1 \), making it easier to apply epsilon-delta rules and find an \( N \) such that \( |f(x) - L| < \epsilon \).
Rational Functions
Rational functions are fractions of polynomial functions, expressed typically as \( \frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomials. They are prevalent in calculus and often analyzed for their behavior as \( x \) approaches certain values or infinity.
A rational function can be simple, like \( \frac{1}{x} \), or more complex like \( \frac{x}{x+1} \), as in our exercise. Analyzing these functions involves looking at their asymptotic behavior, which indicates how a function behaves as it approaches infinity or near points where it might become undefined.
A rational function can be simple, like \( \frac{1}{x} \), or more complex like \( \frac{x}{x+1} \), as in our exercise. Analyzing these functions involves looking at their asymptotic behavior, which indicates how a function behaves as it approaches infinity or near points where it might become undefined.
- For \( \frac{x}{x+1} \), as \( x \rightarrow +\infty \), the function nears \( 1 \).
- Graphically, rational functions can have horizontal asymptotes, vertical asymptotes, or both, depending on the degrees of \( p(x) \) and \( q(x) \).