Question

Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. \(\lim _{x \rightarrow 1 / 10} \frac{1}{x}=10\) (Hint: To find \(\delta,\) you need to bound \(x\) away from 0. So let \(\left.\left|x-\frac{1}{10}\right|<\frac{1}{20} .\right)\)

Short answer

Question: Prove that the limit exists for the function \( f(x) = \frac{1}{x} \) as \( x \) approaches \( \frac{1}{10} \) and find a relationship between \( \epsilon \) and \( \delta \) to guarantee this. Use the hint: "\(|x - \frac{1}{10}| < \frac{1}{20}\)" to bound \(x\) away from 0. Answer: The limit exists for the function \(f(x) = \frac{1}{x}\) as \(x\) approaches \(\frac{1}{10}\), and a relationship between \(\epsilon\) and \(\delta\) that guarantees the existence of the limit is \(\delta = \min{\left\{\frac{1}{20}, \frac{3}{20}\epsilon\right\}}\).

Step-by-step solution

Consider the given limit expression

We are given the limit \(\lim_{x \rightarrow 1 / 10} \frac{1}{x} = 10\). We want to show that for any given \(\epsilon > 0\), we can find a \(\delta > 0\) such that if \(0 < |x - \frac{1}{10}| < \delta\), then \(|\frac{1}{x} - 10| < \epsilon\).

Manipulate the expressions

Let's manipulate \(|\frac{1}{x} - 10|\) to find an expression that can be related to \(\epsilon\) and \(\delta\). We have \(|\frac{1}{x} - 10| = |\frac{1}{x} - \frac{10x}{x}| = |\frac{1 - 10x}{x}| = \frac{|10x - 1|}{|x|}\). Now, we have the inequality: \(|\frac{1}{x} - 10| < \epsilon \Rightarrow \frac{|10x - 1|}{|x|} < \epsilon\).

Use the hint provided to bound \(x\)

We are given the hint to let \(\left|\left|x-\frac{1}{10}\right|<\frac{1}{20}\right.\). Using this, our aim is to bound \(x\) away from 0. Since \(|x - \frac{1}{10}| < \frac{1}{20}\), we get \(-\frac{1}{20} < x - \frac{1}{10} < \frac{1}{20}\). Adding \(\frac{1}{10}\) to all the terms in the inequality, we get \(-\frac{1}{20} + \frac{1}{10} < x < \frac{1}{20} + \frac{1}{10}\), which simplifies to \(\frac{3}{20} < x < \frac{5}{20}\). This bounds \(x\) away from 0.

Relate the expressions for \(\epsilon\) and \(\delta\)

Now, we need to relate the expressions for \(\epsilon\) and \(\delta\). Recall that we have \(\frac{|10x - 1|}{|x|} < \epsilon\). Since \(\frac{3}{20} < x < \frac{5}{20}\), we know that \(|x| > \frac{3}{20}\). Using the hint and the inequality from step 3, redefine the expression as \(\frac{|x -\frac{1}{10}|}{|x|} \cdot |\frac{10x - 1}{x - \frac{1}{10}}| = \frac{|10x - 1|}{|x|} \cdot \frac{|x - \frac{1}{10}|}{|10x - 1|}\). Now, let \(\delta = \min{\left\{\frac{1}{20}, \frac{3}{20}\epsilon\right\}}\). Then \(|\frac{10x - 1|}{|x|} < \frac{|10x - 1|}{\frac{3}{20}}\). Using the definition of \(\delta\), the relation becomes \(\frac{|10x - 1|}{\frac{3}{20}} \cdot \frac{|x - \frac{1}{10}|}{|10x - 1|} < \epsilon\). Since \(|\frac{1}{x} - 10| = \frac{|10x - 1|}{|x|} \cdot \frac{|x - \frac{1}{10}|}{|10x - 1|}\), we have shown that the relation \(|\frac{1}{x} - 10| < \epsilon\) holds if we choose \(\delta = \min{\left\{\frac{1}{20}, \frac{3}{20}\epsilon\right\}}\).

Key concepts

Precise Definition of a Limit

Understanding the precise definition of a limit is critical for mastering calculus. The concept of a limit aims to describe the behavior of a function as the input approaches a particular point. Essentially, a limit answers the question: What value does a function converge to, as its variable gets arbitrarily close to a certain number?

When we state that the limit of a function f(x) as x approaches a value 'a' is L, denoted as \( \lim_{x \rightarrow a} f(x) = L \), what we're asserting is that f(x) can get as close to L as we want provided we choose x sufficiently close to 'a'. What this means is there exists a distance (no matter how small), known as epsilon (\( \epsilon \)), such that for every x within a certain interval around 'a' ─ except 'a' itself ─ the function's value is within epsilon of L. The interval around 'a' is determined by another distance, delta (\( \delta \)), and does not include the point 'a' itself. This definition forms the basis of epsilon-delta proofs, which solidify our understanding of the concept of a limit.

Epsilon-Delta Definition

The epsilon-delta definition provides the foundation for formally proving limits. It employs two arbitrary small positive numbers: epsilon (\( \epsilon \)) and delta (\( \delta \)). Here's how it works:

For a limit \( \lim_{x \rightarrow a} f(x) = L \), we are stating that for every \( \epsilon > 0 \) (however small but greater than zero), there exists a corresponding \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \). In simpler terms, this is saying that the difference between f(x) and L can be made smaller than any positive number \( \epsilon \) by making x close enough to a, within a range defined by \( \delta \) without touching 'a'. This precise relationship between \( \epsilon \) and \( \delta \) forms the crux of limit proofs where we demonstrate this concept for a specific function and target limit value.

Limit of a Function

In the expansive realm of mathematics, the limit of a function is akin to a compass guiding us to the value that a function edges towards as the input gets exceedingly close to a specific point. This concept does not necessarily require the function to be defined at that point or achieve that value; rather, it's about the trend that the function exhibits in the vicinity of the point.

For example, in the step-by-step solution of the limit \( \lim_{x \rightarrow 1/10} \frac{1}{x} = 10 \), the process unraveled a systematic approach to validate how the function \( \frac{1}{x} \) hovers around 10 when x is close to 1/10. The solution utilized the epsilon-delta language, where it fixed a range (delta) around 1/10 and effectively reasoned out how to control the function's output (within epsilon of 10) by keeping x within that delta range. It encapsulates the essence of limits – assessing a function’s behavior at the brink of a specific input value without necessarily reaching it.