Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume \(a\) and \(L\) are finite numbers and assume \(\lim _{x \rightarrow a} f(x)=L\) a. For a given \(\varepsilon>0,\) there is one value of \(\delta>0\) for which \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\) b. The limit \(\lim _{x \rightarrow a} f(x)=L\) means that given an arbitrary \(\delta>0\) we can always find an \(\varepsilon>0\) such that \(|f(x)-L|<\varepsilon\) when ever \(0<|x-a|<\delta\) c. The limit lim \(f(x)=L\) means that for any arbitrary \(\varepsilon>0\) we can always find a \(\delta>0\) such that \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\) d. If \(|x-a|<\delta,\) then \(a-\delta

Short answer

Question: For each of the following statements, identify if it is true or false: a. For a given epsilon, there exists a delta such that if 0 < |x-a| < delta, then |f(x)-L| < epsilon. b. For a given delta, there exists an epsilon such that if 0 < |x-a| < delta, then |f(x)-L| < epsilon. c. For any arbitrary epsilon, there exists a delta such that if 0 < |x-a| < delta, then |f(x)-L| < epsilon. d. If |x - a| < delta, then x is in the open interval (a-delta, a+delta). Answer: a. True b. False c. True d. True

Step-by-step solution

a. True or false?

This statement is true. The given condition is the formal definition of the limit of a function, stating that for every \(\varepsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\varepsilon\). This definition ensures that as \(x\) approaches \(a\), the function's value approaches \(L\), and we can always find an appropriate \(\delta\) for a given \(\varepsilon\).

b. True or false?

This statement is false. The relationship between \(\delta\) and \(\varepsilon\) in the definition of the limit is not arbitrary. In fact, given an \(\varepsilon>0\), we must find a \(\delta>0\) such that \(|f(x)-L|<\varepsilon\) whenever \(0<|x-a|<\delta\). The statement suggests the opposite, which is incorrect.

c. True or false?

This statement is true. It correctly states the formal definition of the limit of a function. For any arbitrary \(\varepsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\varepsilon\). As we get closer to the point \(x=a\), the function's value approaches \(L\).

d. True or false?

This statement is true. If \(|x - a| < \delta\), then \(-\delta < x - a < \delta\). Adding \(a\) to all three parts of the inequality, we obtain \(a-\delta<x<a+\delta\). This means that given the condition \(|x-a|<\delta\), \(x\) will always be in the open interval \((a-\delta, a+\delta)\).

Key concepts

Formal Definition of a Limit

The concept of a limit in calculus is like understanding the behavior of a function as it approaches a certain point. According to the formal definition, we say a function \(f(x)\) has a limit \(L\) as \(x\) approaches \(a\), noted as \(\lim_{x \to a} f(x) = L\), when we can make \(f(x)\) arbitrarily close to \(L\) by choosing \(x\) sufficiently close to \(a\), but not equal to \(a\).

In more formal terms, for every \(\varepsilon > 0\), which represents how close we want \(f(x)\) to be to \(L\), there must be a \(\delta > 0\) such that, for all \(x\) within the distance of \(\delta\) from \(a\), the distance between \(f(x)\) and \(L\) is less than \(\varepsilon\). This idea forms the backbone of many other calculus concepts.

Epsilon-Delta Definition

The epsilon-delta definition of a limit is foundational in calculus. It forms the backbone for rigorously proving limits. Here's how it works:

• \(\varepsilon\) (epsilon) represents any positive number, no matter how small. It's like a window of tolerance around the limit \(L\).
• \(\delta\) (delta) is another positive number, controlling how close \(x\) should be to \(a\).

For a function \(f(x)\) to have a limit \(L\) at \(x = a\), for every \(\varepsilon > 0\), there needs to exist a \(\delta > 0\) such that whenever \(0 < |x - a| < \delta\), then \(|f(x) - L| < \varepsilon\).

This definition gives a precise method to determine if \(f(x)\) gets closer and closer to \(L\) as \(x\) approaches \(a\). It's a bit like using a magnifying glass to zoom in on the behavior of \(f(x)\) near \(a\).

Inequality Manipulation

In mathematics, especially in calculus, manipulating inequalities is crucial when working with limits. One common inequality manipulation seen is when determining how \(|x-a| < \delta\) can imply that \(x\) is within a specific interval.

• If \(|x-a| < \delta\), it means the absolute value of the difference between \(x\) and \(a\) is less than \(\delta\).
• This inequality can be expressed as \(-\delta < x-a < \delta\).
• By adding \(a\) to all parts of the inequality, you get \(a-\delta < x < a+\delta\).

Such manipulations make it easier to visualize how \(x\) is confined to a range or interval around \(a\). It's a simple yet powerful technique that provides insights into the neighborhood around a point.

Continuity of a Function

Continuity of a function relates directly to limits and the epsilon-delta definition. A function \(f(x)\) is said to be continuous at a point \(x = a\) if the following conditions are met:

• \(f(a)\) is defined (the function has a value at \(a\)).
• The limit \(\lim_{x \to a} f(x)\) exists.
• The value of the function at that point equals the limit: \(\lim_{x \to a} f(x) = f(a)\).

This means there are no jumps, breaks, or holes in the graph of the function at \(x = a\).

Continuity ensures that small changes in \(x\) result in small changes in \(f(x)\), maintaining a predictable behavior without sudden shifts. Understanding continuity through limits gives a solid foundation for exploring more advanced calculus topics like derivatives and integrals.