Problem 19 - Calculus for Biology and Medicine

Question

Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow \infty} \frac{x}{x+1}=1 $$

Step-by-Step Solution

Answer

The limit is 1 as shown by the definition of limits, with \( N \) chosen appropriately based on \( \epsilon \).

Step 1: 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 \( \lim_{{x \to \infty}} \frac{x}{x+1} = 1 \).

Step 2: 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: \[\left| \frac{x}{x+1} - 1 \right| = \left| \frac{x - (x+1)}{x+1} \right| = \left| \frac{-1}{x+1} \right| = \frac{1}{x+1}.\]

Step 3: 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:\[ x+1 > \frac{1}{\epsilon} \text{ or equivalently } x > \frac{1}{\epsilon} - 1. \]

Step 4: 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 \( \epsilon > 0 \).

Step 5: 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,\[ \lim_{{x \to \infty}} \frac{x}{x+1} = 1. \]

Key Concepts

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:

• \( \epsilon \) 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 \( |x - a| < \delta \).

In cases involving infinity like \( \lim_{{x \to \infty}} f(x) = L \), the focus is instead on making \( |f(x) - L| < \epsilon \) for all sufficiently large \( x \), because there isn't an actual "point" \( a \) to approach but an idea of "going towards infinity." This establishment of relationship between \( \epsilon \) and the varying value of \( x \) provides a solid proof of limits.

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:

• 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.

Such fertile conceptual terrain of infinity allows calculus to shine, exploring finities and the infinite alike effectively.

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:

• 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 \( x > N \).

These steps beautifully profuse the logical dance of proving with calculus. They rely heavily on algebra, clear argument structuring, and the finesse to relate large-scale problems to manageable components. Calculus proof techniques extend past limit confines, grounding students in persistence, intuition, and procedural skill—essentials in mathematics.