Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that \(\lim _{x \rightarrow c} f(x)=0 \Leftrightarrow \lim _{x \rightarrow c}|f(x)|=0\).

Step-by-Step Solution

Answer

Each condition follows from the other, proving their equivalence.

Step 1: Understanding the Problem

We need to prove that the statement \(\lim _{x \rightarrow c} f(x)=0\) is equivalent to \(\lim _{x \rightarrow c}|f(x)|=0\). This means we have to show both implications: if one side is true, the other must also be true.

Step 2: Proving \(\lim _{x \rightarrow c} f(x)=0\) implies \(\lim _{x \rightarrow c}|f(x)|=0\)

We start by assuming \(\lim _{x \rightarrow c} f(x)=0\). By definition, for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x-c| < \delta\), then \(|f(x) - 0| < \epsilon\). This implies \(|f(x)| < \epsilon\). Since \(|f(x)|\) is already less than \(\epsilon\), the definition of limit shows \(\lim _{x \rightarrow c}|f(x)|=0\).

Step 3: Proving \(\lim _{x \rightarrow c}|f(x)|=0\) implies \(\lim _{x \rightarrow c} f(x)=0\)

Now assume \(\lim _{x \rightarrow c}|f(x)|=0\). This means for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x-c| < \delta\), then \(| |f(x)| - 0| < \epsilon\) or equivalently \(|f(x)| < \epsilon\). With \(|f(x)| < \epsilon\), we also have \(|f(x) - 0| < \epsilon\), which means \(\lim _{x \rightarrow c} f(x)=0\) according to the definition of limit.

Step 4: Conclusion

Since both implications have been shown to be true, we conclude that \(\lim _{x \rightarrow c} f(x)=0 \Leftrightarrow \lim _{x \rightarrow c}|f(x)|=0\). Thus, the equivalence is proven.

Key Concepts

Epsilon-Delta Definition

The epsilon-delta definition is a formal way to define limits in calculus. Limits are fundamental in mathematics to analyze the behavior of functions as they approach a specific point. To better understand this concept, think of two numbers, \(\epsilon\) and \(\delta\).
- \(\epsilon\) (Greek letter epsilon) stands for a very small distance from the limit value.- \(\delta\) (Greek letter delta) is a small distance around the point of interest (in our case, \(c\)).
The definition states: For every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever the distance between \(x\) and \(c\) is less than \(\delta\), the distance between \(f(x)\) and the expected limit value is less than \(\epsilon\).
Put more simply, whatever tiny, desired level of closeness to the limit value you pick (\(\epsilon\)), there’s always a way (by adjusting \(\delta\)) to make sure our function gets that close to the expected value as \(x\) approaches \(c\).
This method of proving limits helps establish rigor and precision, allowing mathematicians to verify that functions behave as expected in calculus.

Absolute Value in Limits

When calculating limits, absolute values can be particularly useful. Absolute value measures the magnitude of a number or expression without considering its sign.
This means that while limiting a function, we worry about the size—how small or large the number is—rather than its position above or below zero.
When you see \(|f(x)|\), it ensures that we focus on the distance between \(f(x)\) and zero, without letting the sign distract us.
In the context of proving limits, - If \(\lim_{x \rightarrow c} |f(x)| = 0\), it means as \(x\) gets closer to \(c\), \(f(x)\) becomes almost zero.- The same idea is true when considering \(\lim_{x \rightarrow c} f(x) = 0\), as their equivalence shows. By using absolute values, we ensure correctness by focusing on distance logic, which is key to maintaining objectivity in limit computations.

Equivalence of Limits

The concept of equivalence in limits involves proving that two statements about limits are logically the same. This is crucial for establishing whether transformations or reinterpretations of a limit statement hold true.
In our example, we need to show that \(\lim_{x \rightarrow c} f(x) = 0\) is equivalent to \(\lim_{x \rightarrow c} |f(x)| = 0\). The implication goes both ways:
- **First Direction**: Suppose \(\lim_{x \rightarrow c} f(x) = 0\). By the epsilon-delta definition, for every \(\epsilon > 0\), there exists \(\delta > 0\) such that \(|f(x)|<\epsilon\). Since absolute value simplifies differences, \(|f(x)|<\epsilon\) is naturally \(|f(x)-0|<\epsilon\), proving \(\lim_{x \rightarrow c} |f(x)| = 0\).
- **Second Direction**: Suppose \(\lim_{x \rightarrow c} |f(x)|=0\). Again, by epsilon-delta logic, for all \(\epsilon > 0\), there is a \(\delta\) making \(|f(x)|<\epsilon\) hold. This naturally means \(|f(x)-0|<\epsilon\) should hold too, confirming \(\lim_{x \rightarrow c} f(x) = 0\).
Through proving these bi-directional implications, we show that these two ways of expressing a limit are indeed interchangeable, reinforcing the accuracy and applicability of the expressions.