Question

Infinite Limits In Exercises 73 and \(74,\) use the \(\varepsilon-\delta\) definition of infinite limits to prove the statement. $$ \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty $$

Short answer

The statement \(\lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty \) is confirmed using the epsilon-delta definition of limits with the relationship \(\delta = \frac{1}{\varepsilon}\).

Step-by-step solution

Understanding the Problem Statement

The objective is to prove that \(\lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty .\) This implies that for any \(\varepsilon > 0 ,\) there exists a \(\delta > 0\) such that if \(0 < x-3 < \delta ,\) then \(\frac{1}{x-3} > \varepsilon . \) The challenge is to find such a delta for a given epsilon.

Deriving the Epsilon-Delta Relationship

As the limit is \(\infty,\) for any number \(\varepsilon > 0,\) there exists a number \(\delta ( = \frac{1}{\varepsilon} )\) such that if \(0 < x-3 < \delta,\) then \( \frac{1}{x-3} > \varepsilon .\) The relation can be re-written as \( x-3 < \delta \implies \frac{1}{x-3} > \varepsilon.\)

Confirmation of the Epsilon-Delta Relationship

To test the validity of the derived relationship, start with \( x-3 < \frac{1}{\varepsilon}.\) Invert both sides, which does not change the inequality since both sides are positive, to get \( \frac{1}{x-3} > \varepsilon .\) Since both expressions match, it can thus be concluded that the statement \(\lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=\infty \) holds using the epsilon-delta definition of limits.

Key concepts

Calculus

Calculus is an essential tool in mathematics that deals with continuous change. At the heart of calculus are the concepts of differentiation and integration, which allow us to analyze the rates at which quantities change and the accumulation of quantities, respectively. One of the pivotal subfields within calculus is the study of limits. A limit describes the behavior of a function as its input approaches a particular point. Limits can be finite, where they approach a specific value, or infinite, where the function's value increases or decreases without bound as it approaches some point.

As students explore the different aspects of calculus, becoming comfortable with the notation and terminology used to describe limits, such as the epsilon-delta definition, is crucial. Grasping these concepts paves the way for understanding more advanced topics, such as continuity, the derivative, and the integral. Furthermore, limits are foundational for many applications, from physics to economics, where relationships must be understood even at points of discontinuity or unbounded behavior.

Infinite Limits

The concept of an infinite limit is encountered when a function's value grows without bound as the input approaches a particular value. Mathematically, we write this as \( \( \lim _{x \rightarrow c} f(x) = \infty \) \), where 'c' is the point that 'x' is approaching.

Understanding infinite limits is necessary because it helps in analyzing the behavior of functions that exhibit extreme values. For instance, in real-world applications, one might be interested in the behavior of a function representing stress on a material as force is applied, where the stress might approach infinity just before the material breaks.

Unlike finite limits, where we say the function approaches a particular value, with infinite limits, we can only describe the behavior in terms of the function growing larger and larger. They are often used in calculus to specifically illustrate the point where a function does not have a finite limit.

Limit Proofs

Limit proofs are a fundamental piece of the puzzle in calculus, particularly when we need to establish the existence of a limit rigorously. The epsilon-delta definition of limits provides a formal method of proving that a certain limit exists.

In the context of the exercise, we used the epsilon-delta approach to prove an infinite limit. For the infinite limit, the epsilon-delta definition is adapted: for every positive number \( \(\varepsilon \) \), no matter how large, there exists a corresponding \( \(\delta \) \), such that for all 'x' within the delta neighborhood around 'c' (but not including 'c'), the value of \(\frac{1}{x-c}\) is greater than \(\varepsilon\). This essentially confirms the function's trend towards infinity as x approaches c.

These proofs are not just academic exercises; they ensure that the theorems and principles we rely on in calculus are built on solid logical foundations. When a student understands how to construct and deconstruct these proofs, they gain a deeper insight into the behavior of functions and the calculus itself.