Question

Use the definition of a limit to prove the following results. \(\lim _{x \rightarrow 1 / 10} \frac{1}{x}=10\) (Hint: To find \(\delta,\) you need to bound \(x\) away from 0. So let \(\left|x-\frac{1}{10}\right|<\frac{1}{20}\).)

Short answer

In summary, we were tasked to prove that the limit \(\lim_{x \rightarrow \frac{1}{10}} \frac{1}{x} = 10\) using the epsilon-delta definition of a limit. Following the steps provided in the solution, we were able to establish a relationship between \(\delta\) and \(\epsilon\): \(\delta(\epsilon)=\min\{\frac{1}{20}, \frac{1}{x(\epsilon)}\}\). This relationship shows that given any \(\epsilon > 0\), we can find some \(\delta\) such that \(0 < \left|x - \frac{1}{10}\right| < \delta\) implies \(|\frac{1}{x} - 10| < \epsilon\). Hence, we have successfully proven that \(\lim_{x \rightarrow \frac{1}{10}} \frac{1}{x} = 10\).

Step-by-step solution

Find a bound for \(x\)

Using the given hint, we will first find a bound for \(x\). We know that \(\left|x - \frac{1}{10}\right| < \frac{1}{20}\). Since we want to show that there exists a bound for \(x\), let's rewrite the inequality as: \(\frac{1}{20} > \left|x - \frac{1}{10}\right| > -\frac{1}{20}\) Now add \(\frac{1}{10}\) to all sides: \(\frac{3}{20} > x > \frac{1}{20}\)

Manipulate the expression given for the limit

Now, we need to manipulate the given limit expression to find a relationship between \(\delta\) and \(\epsilon\). Let's start with the expression: \(|f(x) - L| < \epsilon\) which is equivalent to: \(|\frac{1}{x} - 10| < \epsilon\)

Use the bound of \(x\) to find a relationship

We utilize the bound we found for \(x\) in step 1: \(\frac{3}{20} > x > \frac{1}{20}\) Since \(0 < |x-\frac{1}{10}| < \delta\) implies \(|f(x) - L| < \epsilon\), let's write the inequality with our expression from step 2: \(\frac{1}{20} < \left|x - \frac{1}{10}\right| < \frac{1}{20} \Longrightarrow |\frac{1}{x} - 10| < \epsilon\)

Find \(\delta\) in terms of \(\epsilon\)

Now, we need to find a relationship between \(\delta\) and \(\epsilon\). Let's use the inequalities from step 3 and rewrite them: \(\frac{1}{20} < |\frac{1}{x} - 10| < \frac{1}{20}\) Let's find the \(\delta\) for which the right-hand inequality holds: \(\frac{1}{20} < |\frac{1}{x} - 10|\) Now find \(x\): \(x > \frac{1}{190}\) This gives us a bound for \(\delta\). Hence, we can take \(\delta = \min\{\frac{1}{20}, \frac{1}{x}\}\).

Establish the relationship between \(\delta\) and \(\epsilon\)

Now, going back to step 4, we have: \(\frac{1}{20} < |\frac{1}{x} - 10| < \epsilon\) Note that from step 4 we got that \(x > \frac{1}{190}\), and \(\delta = \min\{\frac{1}{20}, \frac{1}{x}\}\). Therefore, our bound for \(\delta\) depends on \(\epsilon\), and we can use their relationship as: \(\delta(\epsilon)=\min\{\frac{1}{20}, \frac{1}{x(\epsilon)}\}\). So now we have shown that given any \(\epsilon > 0\), we can find some \(\delta\) such that \(0 < \left|x - \frac{1}{10}\right| < \delta\) implies \(|\frac{1}{x} - 10| < \epsilon\). Thus, by the definition of a limit, \(\lim _{x \rightarrow 1 / 10} \frac{1}{x} = 10\) is proven.

Key concepts

Epsilon-Delta Definition of a Limit

Understanding the epsilon-delta definition of a limit is crucial for mastering calculus. This concept serves as the formal foundation for the definition of continuity and limits. To say that \(\lim_{x \rightarrow c}f(x) = L\), we mean for every number \(\epsilon > 0\), there exists a number \(\delta > 0\) such that if \(|x - c| < \delta\) then \(|f(x) - L| < \epsilon\). This ‘dance’ between \(\epsilon\) and \(\delta\) is at the heart of proving limits.

To prove that the limit as \(x\) approaches \(\frac{1}{10}\) of \(\frac{1}{x}\) is 10, one can use an \(\epsilon-\delta\) argument. You start by selecting an \(\epsilon > 0\) arbitrarily. The goal is to find a \(\delta\) based on this \(\epsilon\) such that whenever \(|x - \frac{1}{10}| < \delta\), it guarantees that \(|\frac{1}{x} - 10| < \epsilon\). The proof often involves finding bounds and utilizing inequality manipulation to derive the desired \(\delta\).

Through inequality manipulation, one can deduce the bounds for \(x\) and then tweak these bounds to meet the \(\epsilon-\delta\) criteria, establishing the relationship between \(\delta\) and \(\epsilon\) conclusively. This rigorous approach ensures that the intuitive notion of 'getting closer' is quantified and verifiable.

Inequality Manipulation in Limit Proofs

Inequality manipulation is a powerful tool when it comes to proving limits using epsilon-delta definition. It involves rearranging, adding, or multiplying both sides of an inequality to obtain a new inequality that's easier to work with. The key here is to perform operations that maintain the truth of the original inequality while steering it towards a form that can reveal the \(\delta-\epsilon\) relationship.

In our given problem, we utilize the hint to first set a boundary for \(x\) by stating that \(|x - \frac{1}{10}| < \frac{1}{20}\). Within these steps, we saw inequalities being transformed to find an \(x\) that would satisfy both \(\epsilon\) and \(\delta\) constraints. For instance, flipping the inequality sign appropriately, factoring, and taking reciprocals are all part of the inequality manipulation process.

Keeping the manipulations clear and stepwise allows us to track the changes and ensure that each step logically follows from the last. This process not only leads to the correct \(\delta\) for a given \(\epsilon\) but also strengthens the student's problem-solving skills as they learn to navigate through intricate inequalities with confidence.

Understanding the Limit of a Function

The limit of a function is a fundamental concept in calculus that describes the behavior of a function as its argument gets close to a certain point. It asks the question: as \(x\) approaches some value \(c\), what value does \(f(x)\) approach? The beauty of limits is that they allow us to analyze points where the function might not necessarily be defined, like points of discontinuity or where there's a hole in the graph.

In the exercise, proving the limit \(\lim _{x \rightarrow 1 / 10} \frac{1}{x} = 10\) exemplifies how we can deduce function behavior even at points where direct substitution isn't possible. By establishing that, for any chosen \(\epsilon\), there is a corresponding \(\delta\) that ensures the function's value stays within an \(\epsilon\)-radius of the limit \(L\), we have effectively defined the limit of the function.

Moreover, recognizing the limit concept is essential for advancing in calculus topics such as derivatives and integrals. It helps to know that the delta does not depend on \(x\) but rather on \(\epsilon\) and the conditions we set around the point where \(x\) approaches \(c\). This approach imbues students with the intuition needed to understand more complex concepts that build on the idea of limits.