Problem 39 - Precalculus Mathematics for Calculus

Question

Limits of Sums and Products (a) Show by means of an example that \(\lim _{x \rightarrow a}[f(x)+g(x)]\) may exist even though neither \(\lim _{x \rightarrow a} f(x)\) nor \(\lim _{x \rightarrow a} g(x)\) exists. (b) Show by means of an example that \(\lim _{x \rightarrow a}[f(x) g(x)]\) may exist even though neither \(\lim _{x \rightarrow a} f(x)\) nor \(\lim _{x \rightarrow a} g(x)\) exists.

Step-by-Step Solution

Answer

(a) Limit exists: \(\lim_{x \to 0}[\sin\left(\frac{1}{x}\right) - \sin\left(\frac{1}{x}\right)] = 0\). (b) Limit exists: \(\lim_{x \to 0}[\sin\left(\frac{1}{x}\right) \cdot \frac{1}{x}] = 0\).

Step 1: Understanding the Problem Statement

We are given two tasks to solve with examples: (a) Show that the limit of the sum of two functions can exist even if individual limits do not exist; (b) Show that the limit of the product of two functions can exist even if individual limits do not exist.

Step 1: Choose Example Functions for Part (a)

Consider the functions \(f(x) = \sin\left(\frac{1}{x}\right)\) and \(g(x) = -\sin\left(\frac{1}{x}\right)\). Both of these functions oscillate infinitely as \(x\) approaches 0, so neither \(\lim_{x \to 0} f(x)\) nor \(\lim_{x \to 0} g(x)\) exists.

Step 2: Determine Limit of the Sum for Part (a)

Examine the expression \(f(x) + g(x) = \sin\left(\frac{1}{x}\right) - \sin\left(\frac{1}{x}\right) = 0\). As \(x\) approaches 0, the sum \(f(x) + g(x) = 0\) is constant, hence \(\lim_{x \to 0} [f(x) + g(x)] = 0\).

Step 3: Choose Example Functions for Part (b)

Consider the functions \(f(x) = \sin\left(\frac{1}{x}\right)\) and \(g(x) = \frac{1}{x}\). As \(x\) approaches 0, neither \(\lim_{x \to 0} f(x)\) nor \(\lim_{x \to 0} g(x)\) exists because \(\sin\left(\frac{1}{x}\right)\) oscillates and \(\frac{1}{x}\) tends to infinity.

Step 4: Determine Limit of the Product for Part (b)

Examine \(f(x) \cdot g(x) = \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x}\). As \(x\) approaches 0, this expression behaves like \( \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}} = \frac{\sin(t)}{t} \to 0\) by the Squeeze Theorem, where \(t = \frac{1}{x}\) as \(t\) approaches infinity.

Step 6: Conclusion

In both examples, we have shown that the limit of a sum or product might exist even though the limits of individual functions do not. For part (a), the sum of oscillating functions leads to a constant, and for part (b), the product yields zero due to squashing effects at limits.

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus, often used when finding limits of products where direct substitution does not provide a clear answer. It establishes the limit of a function that is squeezed between two bounding functions that have the same limit at a given point.
For example, in our exercise, we applied the Squeeze Theorem to the product of the functions \( f(x) = \sin\left( \frac{1}{x} \right) \) and \( g(x) = \frac{1}{x} \). As \( x \to 0 \), these individual limits do not exist. However, \( f(x) \cdot g(x) = \sin\left( \frac{1}{x} \right) \cdot \frac{1}{x} \) simplifies to \( \frac{\sin(t)}{t} \) as \( t = \frac{1}{x} \) approaches infinity.
Since \( -1 \leq \sin(t) \leq 1 \), we can multiply through by \( \frac{1}{t} \) (which approaches 0) to squeeze \( \frac{\sin(t)}{t} \) towards 0.
Thus, the Squeeze Theorem helps in determining that the product's limit approaches 0, even if the limits of \( f(x) \) and \( g(x) \) alone don't exist.

Limits of Sums

In calculus, understanding limits of sums can provide surprising results. Even if individual function limits do not exist, their sum can still converge to a limit. This idea is pivotal when dealing with oscillating functions.
For our exercise, consider \( f(x) = \sin\left( \frac{1}{x} \right) \) and \( g(x) = -\sin\left( \frac{1}{x} \right) \). Both of these oscillate infinitely as \( x \to 0 \), leading to non-existent individual limits. However, when these functions are summed, their oscillations perfectly cancel each other out: \( f(x) + g(x) = \sin\left( \frac{1}{x} \right) - \sin\left( \frac{1}{x} \right) = 0 \).
The sum remains constant, and thus, \( \lim_{x \to 0} [f(x) + g(x)] = 0 \).
This demonstrates that strategic pairing of functions in sums can lead to existing limits even when the original functions individually do not have one.

Limits of Products

The concept of limits of products is another intriguing aspect of calculus where a product might converge despite its individual components lacking clear limits. This can lead to unexpected but insightful outcomes for problem-solving.
In the provided exercise, we observe \( f(x) = \sin\left( \frac{1}{x} \right) \) and \( g(x) = \frac{1}{x} \) which, as \( x \to 0 \), individually do not have limits. However, the product, \( f(x)g(x) = \sin\left( \frac{1}{x} \right) \cdot \frac{1}{x} \), utilizes the oscillation of \( \sin\left( \frac{1}{x} \right) \) and the reciprocal \( \frac{1}{x} \) to squash the larger oscillations, effectively leading the product towards \( 0 \).
This showcases how multiplying two non-convergent functions can still exhibit a convergent behavior through interaction. The underlying phenomena often leverage the concepts of bounding and rate binaries within oscillatory motion.

Nonexistent Limits

Nonexistent limits are typically the result of a function's behavior as the input approaches a certain point, but never settling to a finite number. Understanding why limits do not exist can enhance comprehension of complicated functions.
Several factors may contribute to nonexistence:

• The function might oscillate indefinitely, much like \( \sin\left( \frac{1}{x} \right) \) does as \( x \to 0 \), never allowing it to settle.
• Infinity is a result, as seen in \( \frac{1}{x} \) approaching 0. Here, the function grows unbounded.
• Inconsistency from different directions (like from left or right) leads to varied outcomes.

Within the exercise, none of \( f(x) \) or \( g(x) \) had singular limits, illustrating the intriguing facet of mathematics where inputs can refuse conformance to norms. Recognizing when and why limits are nonexistent is crucial for mastering the calculus landscape.