Question

(a) Prove: \(\lim _{x \rightarrow x_{0}} f(x)\) does not exist (finite) if for some \(\epsilon_{0}>0,\) every deleted neighborhood of \(x_{0}\) contains points \(x_{1}\) and \(x_{2}\) such that $$ \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geq \epsilon_{0} $$ (b) Give analogous conditions for the nonexistence of $$ \lim _{x \rightarrow x_{0}+} f(x), \lim _{x \rightarrow x_{0}-} f(x), \quad \lim _{x \rightarrow \infty} f(x), \text { and } \lim _{x \rightarrow-\infty} f(x) $$

Short answer

Question: Prove that if for some given \(\epsilon_{0}>0\), every neighborhood of \(x_{0}\) contains points \(x_{1}\) and \(x_{2}\) such that \(\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \geq \epsilon_{0}\), then \(\lim_{x\rightarrow x_{0}}f(x)\) does not exist. Also, provide analogous conditions for the nonexistence of one-sided limits and limits at infinity. Answer: We proved that if every neighborhood of \(x_0\) contains points \(x_1\) and \(x_2\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\), then the limit \(\lim_{x\rightarrow x_{0}}f(x)\) does not exist. The analogous conditions for the nonexistence of one-sided limits are: (i) For \(\lim_{x\rightarrow x_{0}+}f(x)\): Every neighborhood of \(x_0\) to the right of \(x_0\) contains points \(x_1\) and \(x_2\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\). (ii) For \(\lim_{x\rightarrow x_{0}-}f(x)\): Every neighborhood of \(x_0\) to the left of \(x_0\) contains points \(x_1\) and \(x_2\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\). For limits at infinity: (iii) For \(\lim_{x\rightarrow\infty}f(x)\): For every \(M>0\), there exist points \(x_1\) and \(x_2\) such that \(x_1, x_2 > M\) and \(|f(x_1)-f(x_2)|\geq\epsilon_0\). (iv) For \(\lim_{x\rightarrow-\infty}f(x)\): For every \(M<0\), there exist points \(x_1\) and \(x_2\) such that \(x_1, x_2 < M\) and \(|f(x_1)-f(x_2)|\geq\epsilon_0\).

Step-by-step solution

Show that the condition violates the limit's definition

Let us consider the definition of a limit. If \(\lim_{x\rightarrow x_{0}}f(x)=L\), then for every \(\epsilon>0\), there exists \(\delta>0\) such that if \(0<|x-x_{0}|<\delta\), then \(|f(x)-L|<\epsilon\). However, the given condition states that for every \(\epsilon_0>0\), there exist points \(x_1\) and \(x_2\) in any neighborhood of \(x_0\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\). We will now show that this condition contradicts the definition of a limit.

Prove the contradiction

Let us assume that \(\lim_{x\rightarrow x_{0}}f(x)=L\). Consider any \(\epsilon>0\), such that \(\epsilon = \frac{\epsilon_0}{2}\). According to the limit definition, there exists \(\delta>0\) such that if \(0<|x-x_{0}|<\delta\), then \(|f(x)-L|<\frac{\epsilon_0}{2}\). Now, for this \(\delta\), it is given that there exist points \(x_1\) and \(x_2\) in the neighborhood of \(x_0\), such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\). Now, by the triangle inequality, we have: $$ \left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq \left|f\left(x_{1}\right)-L\right|+\left|f\left(x_{2}\right)-L\right| < \frac{\epsilon_{0}}{2}+\frac{\epsilon_{0}}{2}=\epsilon_{0} $$ This is a contradiction because \(|f(x_1)-f(x_2)| \geq \epsilon_0\), which shows that \(\lim_{x\rightarrow x_{0}}f(x)\) does not exist. (b) Analogs for one-sided and infinite limits:

Analogous conditions for one-sided limits

(i) For \(\lim_{x\rightarrow x_{0}+}f(x)\): Every neighborhood of \(x_0\) to the right of \(x_0\) contains points \(x_1\) and \(x_2\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\). (ii) For \(\lim_{x\rightarrow x_{0}-}f(x)\): Every neighborhood of \(x_0\) to the left of \(x_0\) contains points \(x_1\) and \(x_2\) such that \(|f(x_1)-f(x_2)| \geq \epsilon_0\).

Analogous conditions for limits at infinity

(iii) For \(\lim_{x\rightarrow\infty}f(x)\): For every \(M>0\), there exist points \(x_1\) and \(x_2\) such that \(x_1, x_2 > M\) and \(|f(x_1)-f(x_2)|\geq\epsilon_0\). (iv) For \(\lim_{x\rightarrow-\infty}f(x)\): For every \(M<0\), there exist points \(x_1\) and \(x_2\) such that \(x_1, x_2 < M\) and \(|f(x_1)-f(x_2)|\geq\epsilon_0\).

Key concepts

Definition of a Limit

Understanding the definition of a limit is fundamental in real analysis, as it sets the stage for calculus and deeper mathematical concepts. At its core, a limit tries to describe the behavior of a function as the input approaches a certain point. More formally, we say the limit of function f(x) as x approaches x₀ is L, written as \( \lim \limits_{x \to x_{0}} f(x) = L \), if we can make f(x) as close as we want to L by taking x sufficiently close to x₀.

Mathematically, this translates to the epsilon-delta definition, which is discussed in detail in a later section. The idea is that for every small positive number \( \epsilon \), there's a corresponding small positive number \( \delta \) such that if the distance between x and x₀ is less than \( \delta \), then the distance between f(x) and L will be less than \( \epsilon \). When this condition is violated, as seen in the exercise where the function values differ by at least \( \epsilon_{0} \) within any neighborhood around x₀, the limit does not exist.

One-Sided Limits

One-sided limits investigate the behavior of a function as the input approaches a specific point exclusively from one side — either from the left or the right. For instance, when we consider \( \lim \limits_{x \to x_{0}^{+}} f(x) \), we are interested in the value that f(x) approaches as x gets infinitesimally close to x₀ from the right. Similarly, \( \lim \limits_{x \to x_{0}^{-}} f(x) \) refers to approaching from the left.

The criteria for non-existence of these limits are analogous to that for the overall limit but restricted to one side of the point x₀. If we can find two points, regardless of how close they are to x₀, with function values differing by at least \( \epsilon_{0} \) when approaching from one side, then the one-sided limit does not exist.

Limits at Infinity

Limits at infinity provide us with a way to understand the behavior of functions as the input grows without bound, either positively or negatively. Specifically, we look at \( \lim \limits_{x \to \infty} f(x) \) to find out what value f(x) approaches as x becomes very large, and \( \lim \limits_{x \to -\infty} f(x) \) for when x becomes very negative.

In order for these limits to exist, the values of f(x) need to stabilize to some number as x increases or decreases without bound. The conditions for the non-existence of these limits are similar to those for finite limits, only now we examine how the function behaves far beyond any large magnitude M (positive for \( \infty \) and negative for \( -\infty \)). If we can find two such points where f(x) differs by at least \( \epsilon_{0} \), then the limit at infinity does not exist.

Epsilon-Delta Definition

The epsilon-delta definition is a precise mathematical formulation of the concept of a limit. It provides a rigorous way to prove whether a limit exists and what its value is. This definition states that for a given function f(x) and limit L, the statement \( \lim {x \to x_{0}} f(x) = L \) is true if for every \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that whenever 0 < |x - x_{0}| < \delta, it follows that |f(x) - L| < \epsilon.

Through the exercise provided, we understand that if the conditions of this definition aren't met — for example, if we can find two points within any \delta-neighborhood around x₀ with function values differing by at least some positive \( \epsilon_{0} \) — then the limit does not exist. This definition ensures that the values of f(x) get infinitely close to L as x approaches x₀, which is at the very heart of understanding limits.