Problem 15 - The Calculus with Analytic Geometry

Question

Prove that \(\lim _{x \rightarrow+\infty} f(x)=1\) by applying Definition 4.1.1; that is, for any \(\epsilon>0\), show that there exists a number \(N>0\) such that \(|f(x)-1|<\epsilon\) whenever \(x>N\). $$ f(x)=\frac{x}{x-1} $$

Step-by-Step Solution

Answer

As \( x \to +\infty \), \( f(x) \to 1 \).

Step 1: Understand the given function and the limit

We are given the function \( f(x) = \frac{x}{x-1} \) and need to show that \( \lim_{{x \to +\infty}} f(x) = 1 \) using the formal definition of a limit.

Step 2: Start with the definition of the limit

Recall the definition of the limit: For any \( \epsilon > 0 \), we need to find a number \( N > 0 \) such that \( |f(x) - 1| < \epsilon \) whenever \( x > N \).

Step 3: Manipulate the expression \( |f(x) - 1| \)

Calculate \( |f(x) - 1| \):\( f(x) = \frac{x}{x-1} \Rightarrow f(x) - 1 = \frac{x}{x-1} - 1 = \frac{x - (x-1)}{x-1} = \frac{1}{x-1} \). Thus, \( |f(x) - 1| = \left|\frac{1}{x-1}\right| \).

Step 4: Set \( |f(x) - 1| < \epsilon \)

We want \( |f(x) - 1| = \left|\frac{1}{x-1}\right| < \epsilon \).

Step 5: Solve for \( x \)

Inequality implies \( \left|\frac{1}{x-1}\right| < \epsilon \). This is equivalent to \( \frac{1}{x-1} < \epsilon \).

Step 6: Find \( N \)

Solve \( \frac{1}{x-1} < \epsilon \: \: x - 1 > \frac{1}{\epsilon} \Rightarrow x > \frac{1}{\epsilon} + 1 \). So, let \( N = \frac{1}{\epsilon} + 1 \).

Step 7: Verify the condition

For any \( \epsilon > 0 \), taking \( x > N \) where \( N = \frac{1}{\epsilon} + 1 \), ensures that \( \left|\frac{1}{x-1}\right| < \epsilon \). Hence, \( |f(x) - 1| < \epsilon \).

Key Concepts

Epsilon-Delta Definition

In calculus, the epsilon-delta definition of limits is a formal way of defining the concept of a limit. This definition provides precise criteria for when a function approaches a certain value. The idea is to prove that you can make the function's value as close as you want to the limit by choosing an appropriately large or small input value.

For our exercise, we are given a function and need to show that it approaches 1 as x approaches infinity. Specifically, for any arbitrary small positive number \(\backslash epsilon > 0\), we have to demonstrate that it’s possible to find a corresponding number \(\backslash N > 0\) such that the function remains within a distance of \(\backslash epsilon\) from the limit for all x greater than N.

Here’s how the epsilon-delta definition works:

• You choose any small positive number \(\backslash epsilon\). This represents how close you want to get to the limit.
• You have to find a corresponding positive number \(\backslash N\) so that all values of x greater than N will keep the function's value within an \(\backslash epsilon\) distance from the target limit.

Understanding this concept is vital for mastering calculus and proving limits rigorously.

Limit at Infinity

The limit at infinity describes how a function behaves as its input values grow larger and larger without bound. Essentially, we want to see what value the function is approaching as x heads towards positive or negative infinity.

In our exercise, we need to show that \(\backslash f(x) = \frac{backslash x}{backslash x - 1} \) approaches 1 as x approaches positive infinity. When dealing with limits at infinity, it often helps to simplify the function to see its behavior more clearly.

Observe that for extremely large values of x, the \(\backslash x - 1\) in the denominator is almost the same as x itself. Therefore, we can approximate:

• As x becomes very large, \(\backslash f(x) = \frac{backslash x}{backslash x - 1} \) behaves like \(\backslash frac {backslash x}{backslash x} = backslash 1\).
• Thus, the function \(\backslash f(x)\) gets closer and closer to 1.

By manipulating the function and observing its behavior as x increases, we can firmly establish the limit at infinity.

Algebraic Manipulation

To rigorously prove limits using the epsilon-delta definition, algebraic manipulation is essential. This involves reshaping or transforming the given function to isolate key terms and simplify expressions to make the proof manageable.

For our given function \(\backslash f(x) = \frac{backslash x}{backslash x - 1} \), we started with the expression \(\backslash |backslash f(x) - 1|\) and aimed to show it is less than our epsilon value. Here is how we manipulate the given function:

• First, compute \(\backslash f(x) - 1\: \backslash f(x) = \frac{backslash x}{backslash x - 1} \)
• Subtract 1 to get \(\backslash |backslash f(x) - 1|\).
• Simplify the result: \(\backslash frac{backslash x}{backslash x - 1} - 1 = backslash frac{backslash x - (backslash x - 1)}{backslash x - 1} = backslash frac{1}{backslash x - 1}\).

Through algebraic manipulation, simplify this expression and set it less than an arbitrary \(\backslash epsilon\): \(\backslash Fair {backslash frac{1}{backslash x - 1} < backslash epsilon \). Solving this inequality for x gives: \ backslash x > \backslash frac{1}{backslash epsilon} + 1 \.

By identifying the appropriate \(\backslash N\) and verifying that x values greater than \(\backslash N\) satisfy the condition, we complete the proof. Therefore, algebraic manipulation is a powerful tool to make formal limit proofs accessible and manageable.