Chapter 3: Problem 19
Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow \infty} \frac{x}{x+1}=1 $$
Short Answer
Expert verified
The limit is 1 as shown by the definition of limits, with \( N \) chosen appropriately based on \( \epsilon \).
Step by step solution
01
Understanding the Limit Definition
The formal definition of limits states that for \( \lim_{{x \to \infty}} f(x) = L \), for every \( \epsilon > 0 \), there exists \( N > 0 \) such that if \( x > N \), then \( |f(x) - L| < \epsilon \). Here, we want to show that \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \).
02
Finding the Expression
We need to consider the absolute value \( \left| \frac{x}{x+1} - 1 \right| < \epsilon \). Simplifying the expression inside the absolute value: \[\left| \frac{x}{x+1} - 1 \right| = \left| \frac{x - (x+1)}{x+1} \right| = \left| \frac{-1}{x+1} \right| = \frac{1}{x+1}.\]
03
Setting the Inequality
According to the epsilon-definition of limits, we want \( \frac{1}{x+1} < \epsilon \). To satisfy this inequality, we need an \( x \) such that:\[ x+1 > \frac{1}{\epsilon} \text{ or equivalently } x > \frac{1}{\epsilon} - 1. \]
04
Determining the Value of N
By the above, if we take \( N = \frac{1}{\epsilon} - 1 \), whenever \( x > N \), the condition \( \frac{1}{x+1} < \epsilon \) will be satisfied. Thus, the inequality holds for any \( \epsilon > 0 \).
05
Conclusion
We have shown that for any positive \( \epsilon \), we can choose \( N = \frac{1}{\epsilon} - 1 \) such that whenever \( x > N \), \( \left| \frac{x}{x+1} - 1 \right| < \epsilon \). Therefore, by the formal definition of limits,\[ \lim_{{x \to \infty}} \frac{x}{x+1} = 1. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Epsilon-Delta Definition
The epsilon-delta definition is a key concept in understanding limits in calculus. It lays the groundwork for proving limits rigorously. This definition states that for a function \( f(x) \) approaching a limit \( L \) as \( x \to a \), for every \( \epsilon > 0 \), there is a \( \delta > 0 \) such that if the distance between \( x \) and \( a \) (i.e., \( |x - a| \)) is less than \( \delta \), then the function values \( f(x) \) are within \( \epsilon \) of \( L \) (i.e., \( |f(x) - L| < \epsilon \)).
Let's break it down:
Let's break it down:
- \( \epsilon \) represents how close \( f(x) \) gets to \( L \).
- \( \delta \) represents how close \( x \) gets to \( a \).
- The goal is to find a \( \delta \) for every possible \( \epsilon \) that makes \( |f(x) - L| < \epsilon \) true whenever \( |x - a| < \delta \).
Infinity in Limits
When working with limits as \( x \to \infty \), it is important to rethink the standard approach to limits. Unlike the typical \( a \to b \) approach, we consider how values behave as \( x \) grows larger without bounds.
In essence, we ask how a function behaves when \( x \) becomes extremely large. For instance, if you're tasked with finding \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \), you are interested in how \( \frac{x}{x+1} \) approaches 1 as the denominator and numerator values explode into vast numbers.
Visualize this by imagining a large value of \( x \). If \( x = 1000 \), \( \frac{x}{x+1} = \frac{1000}{1001} \approx 0.999 \), showing it approaches 1. What is key with \( \lim_{{x \to \infty}} \) is this "chase" toward the limit value, which is approached but never exceeded. Through this understanding:
In essence, we ask how a function behaves when \( x \) becomes extremely large. For instance, if you're tasked with finding \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \), you are interested in how \( \frac{x}{x+1} \) approaches 1 as the denominator and numerator values explode into vast numbers.
Visualize this by imagining a large value of \( x \). If \( x = 1000 \), \( \frac{x}{x+1} = \frac{1000}{1001} \approx 0.999 \), showing it approaches 1. What is key with \( \lim_{{x \to \infty}} \) is this "chase" toward the limit value, which is approached but never exceeded. Through this understanding:
- Recognize the distinct absence of a "target point" since infinity isn't a number.
- Deduce that as \( x \to \infty \), we only need function closeness towards a certain value.
- The inequality becomes \( |f(x) - L| < \epsilon \), applicable at disappearingly smaller differences from \( L \) as \( x \) grows.
Calculus Proof Techniques
The elegance of calculus lies in its proof techniques, showcasing both creativity and rigor. In solving limits, particularly when using epsilon-delta approaches, several proof strategies are often employed.
For instance, in proving \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \), the strategy involves:
For instance, in proving \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \), the strategy involves:
- Expressing the limit setup using the epsilon-delta format to outline the goal as \( |f(x) - L| < \epsilon \).
- Simplifying the function expression: Transform \( |\frac{x}{x+1} - 1| \) to \( |\frac{-1}{x+1}| \). This represents the core simplification step in the solution.
- Setting an inequality: You then outline \( \frac{1}{x+1} < \epsilon \), restating the problem in terms of epsilon.
- Determining \( N \): Work out \( N = \frac{1}{\epsilon} - 1 \) so it becomes applicable for any \( \epsilon > 0 \) with \( x > N \).