Problem 62 - Calculus: Early Transcendentals

Question

Show by means of an example that \( \displaystyle \lim_{x \to a}\left[ f(x) + g(x) \right] \) may exist even though neither \( \displaystyle \lim_{x \to a}f(x) \) nor \( \displaystyle \lim_{x \to a}g(x) \) exists.

Step-by-Step Solution

Answer

The limit of the sum is 0, even though individual limits do not exist.

Step 1: Understand the Limitation

We are trying to show a case where the limit of a sum exists even if the individual limits do not. This means the expressions for \( f(x) \) and \( g(x) \) must be such that as \( x \) approaches \( a \), both are undefined or do not stabilize to a particular value.

Step 2: Choose Functions f(x) and g(x)

Choose the functions as \( f(x) = \sin\left(\frac{1}{x-a}\right) \) and \( g(x) = -\sin\left(\frac{1}{x-a}\right) \). Both functions oscillate infinitely between -1 and 1 as \( x \to a \), hence their individual limits do not exist.

Step 3: Calculate the Sum f(x) + g(x)

The sum of the functions is: \[ f(x) + g(x) = \sin\left(\frac{1}{x-a}\right) + (-\sin\left(\frac{1}{x-a}\right)) = 0. \]

Step 4: Evaluate the Limit of the Sum

Since \( f(x) + g(x) = 0 \) for all \( x \), we have: \[ \lim_{x \to a} [ f(x) + g(x) ] = \lim_{x \to a} 0 = 0. \] Hence, this limit exists.

Key Concepts

Oscillating Functions

Oscillating functions are those that do not settle at a single value as their input, in this case, the variable approaches a specific number. Instead, they continually vary, often fluctuating between two bounds.
A classic example of an oscillating function is the sine function. When we consider functions like \( f(x) = \sin(\frac{1}{x-a}) \), as \( x \to a \), the argument of the sine function goes to infinity, causing the sine to oscillate continuously between -1 and 1. This means that \( f(x) \) doesn’t stabilize at any single value as \( x \to a \), thus the limit of \( f(x) \) does not exist.
Oscillating functions are crucial in mathematical analysis as they help illustrate scenarios where conventional limits fail to exist individually, but under certain operations, like addition, may yield a stable limit.

Existence of Limit

The existence of a limit indicates whether a function approaches a specific value as its input approaches some point. For a limit \( \lim_{x \to a} f(x) \) to exist, \( f(x) \) must get arbitrarily close to a particular value as \( x \) gets nearer to \( a \).
However, some functions, especially oscillating ones, do not meet this criterion on their own due to their infinite oscillations. Still, it is possible for a combination of such functions to have a limit.
When working with the sum \( f(x) + g(x) \), even if \( f(x) \) and \( g(x) \) do not have individual limits as they approach \( a \) due to oscillations, the behavior of their sum can be different. For example, \( f(x) = \sin(\frac{1}{x-a}) \) and \( g(x) = -\sin(\frac{1}{x-a}) \) each oscillate infinitely, but their sum is consistently zero regardless of \( x \). Hence, this sum has a limit as \( x \to a \).

Sum of Functions

The sum of functions is a fundamental operation where each function value is added together for corresponding inputs. Understanding how limits behave under the sum operation is key in calculus.
In the problem given, we see an illustrative case where despite neither \( \lim_{x \to a} f(x) \) nor \( \lim_{x \to a} g(x) \) existing individually, the sum \( \lim_{x \to a} [f(x) + g(x)] \) does exist. By choosing \( f(x) = \sin(\frac{1}{x-a}) \) and \( g(x) = -\sin(\frac{1}{x-a}) \), each oscillates but in exact opposite directions, leading their sum to be zero for all \( x \).
This shows that under certain conditions, particularly where functions are constructed to cancel their oscillations out, a stable limit for their sum can exist, highlighting an intriguing aspect of function behavior and plot interactions in mathematical analysis.