Problem 37 - Precalculus: Mathematical 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 of sum exists (0). (b) Limit of product exists (0).

Step 1: Understand the Problem

We are given two separate tasks, both involving the use of limits in calculus. In part (a), we need to show an example where the limit of a sum of two functions exists even though the limits of the individual functions do not. In part (b), we need to show a similar example but for the product of two functions.

Step 2: Choose an Example for Part (a)

Let's consider the functions: 1. \( f(x) = \sin(\frac{1}{x}) \) and 2. \( g(x) = -\sin(\frac{1}{x}) \).Neither \( \lim _{x \rightarrow 0} f(x) \) nor \( \lim _{x \rightarrow 0} g(x) \) exists because \( \sin(\frac{1}{x}) \) oscillates between -1 and 1 as \( x \) approaches zero. However, their sum \( f(x) + g(x) = 0 \) which is a constant.

Step 3: Calculate the Limit for Part (a)

Since \( f(x) + g(x) = 0 \) for all \( x \), the limit \( \lim _{x \rightarrow 0}[f(x) + g(x)] = 0 \) does exist because a constant function is continuous.

Step 4: Choose an Example for Part (b)

Consider the functions: 1. \( f(x) = \sin(\frac{1}{x}) \) and 2. \( g(x) = \frac{1}{x} \).As in part (a), \( \lim _{x \rightarrow 0} f(x) \) does not exist due to oscillation, and \( \lim _{x \rightarrow 0} g(x) = \pm \infty \) does not exist because \( \frac{1}{x} \) becomes unbounded.

Step 5: Calculate the Limit for Part (b)

The product \( f(x)g(x) = \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x} \).As \( x \to 0, \sin\left(\frac{1}{x}\right) \) remains bounded between -1 and 1, while \( \frac{1}{x} \to \pm \infty \). But crucially, \( f(x)g(x) \) behaves similarly to the limit of \( \frac{\sin\left(\frac{1}{x}\right)}{x} \) as \( x \to 0 \), which approaches zero by the Squeeze Theorem since \( \frac{-1}{x} \leq \frac{\sin(\frac{1}{x})}{x} \leq \frac{1}{x} \). Hence, \( \lim _{x \rightarrow 0}[f(x)g(x)] = 0 \).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus for finding limits of functions that might be difficult to evaluate directly. It is especially useful when dealing with oscillating functions or as in our example when trying to evaluate a limit of a product where individual limits do not exist.

• The Squeeze Theorem operates by taking advantage of bounding inequalities.
• If a function is sandwiched between two other functions that have the same limit at a point, then it must also converge to that same limit.

In the example presented, we had a product consisting of the oscillating function \( \sin\left(\frac{1}{x}\right) \) multiplied by \( \frac{1}{x} \). As \( x \to 0 \), the product itself oscillates. However, using the inequality:\[ \frac{-1}{x} \leq \frac{\sin(\frac{1}{x})}{x} \leq \frac{1}{x} \]It can be concluded with the Squeeze Theorem that:\[ \lim_{x \to 0} \frac{\sin(\frac{1}{x})}{x} = 0 \]Hence, even though \( \lim_{x \to 0} \sin(\frac{1}{x}) \) and \( \lim_{x \to 0} \frac{1}{x} \) do not independently exist, the limit of their product does, affirming the usefulness of the Squeeze Theorem.

Oscillating Functions

Oscillating functions are those that do not settle on a single value as they approach a particular point. This characteristic often leads to limits that do not exist at that specific point. A perfect example of an oscillating function is \( \sin\left(\frac{1}{x}\right) \). As \( x \) approaches zero, the values of \( \frac{1}{x} \) can become very large, causing the sine function to oscillate wildly between -1 and 1.

• This rapid fluctuation is the main reason why the limit at \( x = 0 \) does not exist for \( \sin\left(\frac{1}{x}\right) \).
• For students, understanding oscillation's impact on limits is crucial, as it helps identify scenarios where limits might not exist.

Moreover, despite this non-existence of particular limits, combinations with other functions, such as those shown in the original problem, can yield existing limits upon summation or multiplication.

Nonexistent Limits

In calculus, a limit might not exist due to several reasons, and it's fundamental to recognize these scenarios to prevent confusion. From the examples given, we saw two main causes of nonexistent limits:

• **Oscillation:** When a function swings back and forth without approaching a specific value, like \( \sin(\frac{1}{x}) \) near \( x = 0 \).
• **Unbounded Behavior:** Functions like \( \frac{1}{x} \) which become infinitely large as \( x \to 0 \).

Nonexistent limits can seem daunting, but understanding them helps in solving complex calculus problems where adding or multiplying these troublesome functions provide insights. Even if the individual limits of functions \( f(x) \) and \( g(x) \) do not exist, certain arithmetic operations could result in a viable limit. For example, the sum \( f(x) + g(x) = 0 \) exists simply because it is a constant function overall despite each function's individual limit not existing.