Problem 20 - 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 proven using the definition: as \( x \to -\infty \), \( \frac{x}{x+1} \to 1 \).

Step 1: Understand the Limit Definition

The formal definition of a limit, in this context, involves proving that for every small positive number \( \varepsilon \), there exists a real number \( N \) such that if \( x < N \), then the inequality \( \left| \frac{x}{x+1} - 1 \right| < \varepsilon \) holds. This will show that as \( x \) approaches \(-\infty\), the function approaches 1.

Step 2: Simplify the Expression

First, simplify the expression \( \frac{x}{x+1} - 1 \).\[\frac{x}{x+1} - 1 = \frac{x}{x+1} - \frac{x+1}{x+1} = \frac{x - (x+1)}{x+1} = \frac{-1}{x+1}\]

Step 3: Apply the Epsilon Condition

We need \( \left| \frac{-1}{x+1} \right| < \varepsilon \). This simplifies to finding \( |x+1| > \frac{1}{\varepsilon} \). To ensure \( x \) is sufficiently large and negative, this condition simplifies to ensuring \( x < -1 - \frac{1}{\varepsilon} \).

Step 4: Choose the Appropriate \( N \)

Set \( N = -1 - \frac{1}{\varepsilon} \). This choice satisfies the requirement that if \( x < N \), then \( |x+1| > \frac{1}{\varepsilon} \) and consequently \( \left| \frac{-1}{x+1} \right| < \varepsilon \).

Step 5: Conclusion

Since for any \( \varepsilon > 0 \), we have found an \( N \) such that for all \( x < N \), \( \left| \frac{-1}{x+1} \right| < \varepsilon \), the formal definition of the limit is satisfied, proving that \( \lim_{x \to -\infty} \frac{x}{x+1} = 1 \).

Key Concepts

Epsilon-Delta Definition

The epsilon-delta (\( \varepsilon \)-\( \delta \)) definition is a cornerstone concept in calculus used to rigorously define what a limit is. It’s typically utilized to confirm a suspected limit by making precise what it means for a function to get ‘close’ to a particular value as the input approaches some value. In the case of a function approaching a limit as \( x \) goes to \( -\infty \), we're concerned with making the function values close to the limit by ensuring the input \( x \) is sufficiently far in the negative direction. The \( \varepsilon \) represents any small positive number indicating closeness.We must find an \( N \) such that when \( x < N \), the absolute difference between the function and the limit is less than the chosen \( \varepsilon \). Ensuring this helps prove that as the input moves sufficiently towards \( -\infty \), the function value increasingly approaches the limit. This exercise allowed us to dive into such reasoning by proving \( \lim_{x \rightarrow -\infty} \frac{x}{x+1} = 1 \).

Limits at Infinity

"Limits at infinity" specifically refer to understanding what happens to a function's value as the input variable grows larger and larger in the positive or negative direction. These limits help describe the end behavior of functions. For \( \lim_{x \rightarrow -\infty} \frac{x}{x+1} = 1 \), the focus is on what the quotient \( \frac{x}{x+1} \) nears as x becomes negatively large. Essentially, as \( x \) decreases beyond all bounds, the expression \( x+1 \) in the denominator becomes much like \( x \) itself, simplifying the function to approximately \( \frac{x}{x} \) or 1. This section of the exercise serves as a showcase for looking at limits as x approaches infinitely large positive or negative values. It’s important to note the distinction between this and limits that approach specific finite values.

Formal Limit Proof

In calculus, a formal limit proof follows a structured approach using the \( \varepsilon \)-\( N \) definition to show definitively that a limit is reached under specific conditions. For \( \lim_{x \rightarrow -\infty} \frac{x}{x+1} = 1 \), constructing a formal proof requires:

• Simplifying the expression to find \( \frac{x}{x+1} - 1 \), which results in \( \frac{-1}{x+1} \) after simplification.
• Applying the condition \( \left| \frac{-1}{x+1} \right| < \varepsilon \) to notify closeness.
• Determining an appropriate \( N \) value where \( N = -1 - \frac{1}{\varepsilon} \), which ensures that for all \( x < N \), the inequality holds true.

This process confirms that no matter how small \( \varepsilon \) is, a corresponding threshold \( N \) can be found such that the limit holds firm as \( x \) continues its way to negative infinity. This methodology is not only vital for proving limits, but it also builds foundational understanding for more complex calculus concepts.