Problem 54 - Thomas Calculus

Question

Another wrong statement about limits Show by example that the following statement is wrong. $$ \begin{array}{l}{\text { The number } L \text { is the limit of } f(x) \text { as } x \text { approaches } c \text { if, given any }} \\ {\epsilon>0, \text { there exists a value of } x \text { for which }|f(x)-L|<\epsilon \text { . }}\end{array} $$ Explain why the function in your example does not have the given value of \(L\) as a limit as \(x \rightarrow c .\)

Step-by-Step Solution

Answer

The statement is wrong because it allows a single \( x \) rather than all \( x \) near \( c \), violating the proper definition of a limit.

Step 1: Understanding the Statement

The statement asserts that for a number \( L \) to be the limit of \( f(x) \) as \( x \) approaches \( c \), there must exist a value of \( x \) such that the condition \(|f(x) - L| < \epsilon\) holds for any \( \epsilon > 0 \). This suggests just one value of \( x \) is sufficient, which is not correct.

Step 2: Correct Definition of Limit

The correct definition requires that for every \( \epsilon > 0 \), there must exist a \( \delta > 0 \) such that for all \( x \) satisfying \( 0 < |x - c| < \delta \), the condition \( |f(x) - L| < \epsilon \) holds. It is important that this holds for all values of \( x \) sufficiently close to \( c \), not just one.

Step 3: Choosing a Function

Consider the function \( f(x) = 1 \) for all \( x eq 0 \) and \( f(0) = 0 \). We want to test if \( L = 1 \) can be a limit as \( x \rightarrow 0 \).

Step 4: Checking the Faulty Condition

According to the faulty definition, for \( \epsilon = 0.5 \), there is a value of \( x \), say \( x = 0.1 \), such that \( |f(x) - 1| < 0.5 \). This is true for \( f(0.1) = 1 \).

Step 5: Applying the Correct Definition

Using the correct definition, we should find a \( \delta \) such that for all \( x \) with \( 0 < |x| < \delta \), \( |f(x) - 1| < \epsilon \) holds. However, \( f(x) = 1 \) already, and only \( f(0) \) is different, i.e., 0, which means \( |f(0) - 1| = 1 \geq 0.5 \). Thus, the \( \delta \) condition fails.

Step 6: Conclusion

While the faulty definition is satisfied because there exists at least one \( x \) (e.g. \( x = 0.1 \)) where \( |f(x) - 1| < \epsilon \), it does not hold for all necessary values around 0. Therefore, \( L = 1 \) is not the limit by the correct definition.

Key Concepts

Epsilon-Delta Definition

In calculus, the concept of a limit is foundational. To fully grasp limits, it's essential to understand the rigorous definition that employs epsilon (\(\epsilon\)) and delta (\(\delta\)). This definition ensures that a function \(f(x)\) approaches the limit \(L\) as \(x\) approaches \(c\). For every small number \(\epsilon > 0\), however tiny we choose it, we can find a corresponding number \(\delta > 0\). This means that whenever \(x\) gets close to \(c\) such that \(0 < |x-c| < \delta\), the function's value \(f(x)\) becomes as close as we wish to \(L\) so that \(|f(x) - L| < \epsilon\).

• The epsilon-delta definition is precise and not just an approximation.
• It's important because it doesn't rely on specific points but an entire interval around the point of interest.
• This ensures uniform closeness of function values to the limit.

Consider it like being at the center of a spotlight: wherever you move within that spotlight, you remain under it, making it fully consistent. To prove limits, demonstrating this definition holds true for every selected \(\epsilon\) is crucial.

Limit Misconceptions

Misconceptions regarding limits are quite common but important to clarify. One major misconception is thinking that if a function gets close to a number at some single point, then it is the limit. This oversimplification often leads to incorrect conclusions.

• Limits concern the behavior of a function as it approaches a point, not only at a single point.
• You need all the surrounding values within a particular radius to exhibit consistent behavior, not just one favorable instance.
• This is why the epsilon-delta formalism is critical—it provides a failsafe against such misconceptions.

Using an incorrect definition, like stating a limit exists if there's only one value of \(x\) for which \(|f(x) - L| < \epsilon\), can mislead. Such a condition might hold for one isolated \(x\) (a fluke), but fail to maintain consistency across a \(\delta\) neighborhood. This confusion was apparent in our example exercise where the faulty definition implies existence if only one suitable \(x\) is found.

Functional Limit Examples

To better internalize the concept of limits, it helps to explore examples. Let's consider the example from our exercise: a function \(f(x)\) defined as \(f(x) = 1\) where \(x eq 0\) and \(f(0) = 0\). The question is whether \(L = 1\) is the limit as \(x\) approaches 0.

• Using the faulty limit assumption, you might think \(L = 1\) is valid because for some \(x\), e.g., \(x=0.1\), \(|f(x) - 1| < \epsilon\) holds true.
• However, applying the correct epsilon-delta criterion reveals inconsistencies, notably at \(x=0\)
• Here, \(|f(0) - 1| = 1\) which does not satisfy the closeness requirement, proving \(L = 1\) can't be the limit.

In practice, examples like this showcase why a limit requires surrounding support, not just isolated achievements. A similar understanding can be built with other examples, where applying the epsilon-delta definition helps prevent falling into traps from misleading observations.