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. \(\lim _{x \rightarrow 3} \frac{1}{x}=\frac{1}{3}(\text {Hint: As } x \rightarrow 3,\) eventually the distance between \(x\) and 3 is less than \(1 .\) Start by assuming \(|x-3|<1\) and show \(\frac{1}{|x|}<\frac{1}{2}\)

Short answer

Short Answer: To prove the limit \(\lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3}\), we use the Epsilon-Delta definition of the limit. We find that by choosing \(\delta= \frac{3}{2}\varepsilon\), the condition \(\left|\frac{1}{x} - \frac{1}{3}\right| < \varepsilon\) is satisfied whenever \(0 < |x-3| < \delta\). Thus, we can conclude that the limit \(\lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3}\).

Step-by-step solution

Understand the Epsilon-Delta definition of a limit

Given a function \(f(x)\), and a point \(x=a\), if for every number \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that $$ |f(x)-L|<\varepsilon $$ whenever $$ 0<|x-a|<\delta $$ then the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\) and we write $$ \lim _{x \rightarrow a} f(x)=L $$

Define the function and limit

We are given the following limit to prove: $$ \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} $$ To prove this limit, we need to find a relationship between \(\varepsilon\) and \(\delta\) which guarantees the existence of the limit.

Assume \(|x-3|

Let's start by assuming \(|x-3|<1\), which means the distance between \(x\) and \(3\) is less than \(1\). Now we want to show that this implies \(\frac{1}{|x|}<\frac{1}{2}\). Since we know \(|x-3|<1\), this means \(-1<x-3<1\) or \(2<x<4\). Thus, \(2<|x|<4\). Reciprocates these inequalities, we obtain: $$ \frac{1}{4}<\frac{1}{|x|}<\frac{1}{2} $$

Find relationship between \(\varepsilon\) and \(\delta\)

Now we want to find a relationship between \(\varepsilon\) and \(\delta\) that guarantees $$ \left|\frac{1}{x} -\frac{1}{3}\right| < \varepsilon $$ whenever $$ 0 < |x-3| < \delta $$ First, note that $$ \left|\frac{1}{x} -\frac{1}{3}\right| = \left|\frac{3-x}{3x}\right| = \frac{|x-3|}{3|x|} $$ We want to make \(\frac{|x-3|}{3|x|}<\varepsilon\). From step 3, we know \(\frac{1}{|x|}<\frac{1}{2}\). Thus, $$ \frac{|x-3|}{3|x|} < \frac{|x-3|}{3\cdot\frac{1}{2}} = \frac{2}{3}|x-3| $$ Now, we can choose \(\delta\) such that $$ \frac{2}{3}|x-3| < \varepsilon $$ Thus, we can choose $$ \delta = \frac{3}{2}\varepsilon $$

Conclusion

By choosing \(\delta= \frac{3}{2}\varepsilon\), we have: $$ \left|\frac{1}{x} -\frac{1}{3}\right| = \frac{|x-3|}{3|x|} < \frac{2}{3}|x-3| < \varepsilon $$ whenever $$ 0 < |x-3| < \delta $$ Therefore, we have proved the limit: $$ \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} $$

Key concepts

Epsilon-Delta Definition

The Epsilon-Delta definition is a foundational concept in calculus, used to rigorously define the limit of a function. Simply put, this definition ensures that as our input value gets arbitrarily close to a specific point, the output value of the function gets arbitrarily close to a desired limit.To understand this deeply, consider the function \( f(x) \). If \( x \) approaches a point \( a \), and for every tiniest positive number \( \varepsilon \) (epsilon), there is a corresponding small positive number \( \delta \) (delta), maintaining the inequality \(|f(x) - L| < \varepsilon\), provided that \(0 < |x - a| < \delta\), then the function \( f(x) \) has a limit \( L \) as \( x \rightarrow a \). This definition is crucial because it establishes a precise language for talking about limits, making concepts in calculus both formal and robust.

Limit Proofs

Limit proofs involve verifying that a function approaches a specific value as its input approaches a particular point. Using the Epsilon-Delta definition makes these proofs rigorous.To prove the limit of a function, you need to:

• Define the given limit problem, specifying the function and the point of interest.
• Identify \( \varepsilon \) and find a matching \( \delta \) that works for the specific function and point.
• Show that the calculated \( \delta \) aligns with the constraints, ensuring \(|f(x) - L| < \varepsilon\).

In our example of proving \( \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} \), these steps ensure we consider all possible cases and confirm the function behaves consistently as \( x \) approaches 3. The choice of \( \delta = \frac{3}{2}\varepsilon \) helps tailor the function's closeness to the limit, reinforcing the theorem's desired outcome.

Reciprocals

Reciprocals, a fundamental concept in mathematics, refer to the inverse of a number. Given a non-zero number \( a \), its reciprocal is \( \frac{1}{a} \). This inverse maintains the property that multiplying a number by its reciprocal yields 1: \( a \times \frac{1}{a} = 1 \).In the context of limit proofs, especially when dealing with fractional functions like \( \frac{1}{x} \), understanding reciprocals becomes crucial. When manipulating inequalities, calculating reciprocals helps determine bounds and assists in establishing relationships between variables. For instance, knowing that as \( x \to 3 \), \( \frac{1}{|x|} < \frac{1}{2} \) helps set tight constraints for our limits proof. This concept allows you to find and verify boundaries, ensuring the correctness of the solution.

Inequality Manipulation

Inequality manipulation is a critical skill in solving calculus problems, particularly when working with limits. Handling inequalities allows us to refine our understanding of relationships between terms and to set boundaries that ensure solutions meet necessary conditions.When working on limit proofs, such as proving \( \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} \), manipulating inequalities involves:

• Assuming initial conditions, such as \( |x - 3| < 1 \), breaking down the absolute value terms for clarity.
• Transforming inequalities to make terms comparable, such as converting \( -1 < x - 3 < 1 \) into \( 2 < x < 4 \).
• Reversing or reciprocating inequalities appropriately, like determining \( \frac{1}{4} < \frac{1}{|x|} < \frac{1}{2} \).

These steps are pivotal in shaping the proofs into precise arguments that demonstrate how a given condition leads ultimately to the desired limit, confirming the theorem's truthfulness and expanding your understanding of the function's behavior.