Limits: Existence and Non-Existence

When we say the limit of a function \( f(x) \) exists at a point \( x = a \), we mean that as \( x \) approaches \( a \), the function \( f(x) \) approaches a specific value \( L \). Mathematically, this is written as:

\[ \lim_{x \to a} f(x) = L \]

This means that for any small value \( \epsilon \), there exists a small enough \( \delta \) such that for all \( x \) within \( \delta \) of \( a \), \( f(x) \) is within \( \epsilon \) of \( L \). The function approaches \( L \) as \( x \) gets closer to \( a \). The key idea is that \( f(x) \) gets arbitrarily close to \( L \) as \( x \) approaches \( a \).

If the limit exists, we say the function is continuous at that point, meaning that there are no jumps or breaks in the graph of the function at \( x = a \).

Consider the function \( f(x) = 2x + 3 \). Let’s find the limit as \( x \to 1 \):

\[ \lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5 \]

As \( x \) approaches 1, the function value approaches 5, and we can see that the limit exists and is equal to 5.

The graph of this function is a straight line, so there is no discontinuity, and the limit exists everywhere for this function.

Let’s consider the function \( f(x) = \frac{1}{x} \) and find the limit as \( x \to 2 \):

\[ \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} \]

As \( x \) approaches 2, the function approaches \( \frac{1}{2} \), so the limit exists and is equal to \( \frac{1}{2} \).

When we say that the limit of a function does not exist at a point \( x = a \), it means that the function does not approach a single value as \( x \) approaches \( a \). There are several reasons why the limit might not exist:

• Discontinuity: If there is a jump or break in the graph of the function at \( x = a \), the limit may not exist.
• Unbounded Behavior: If the function approaches infinity or negative infinity as \( x \) approaches \( a \), the limit does not exist.
• Oscillatory Behavior: If the function oscillates (like sine and cosine) as \( x \) approaches \( a \), the limit does not exist because the function does not settle on a single value.

Consider the following piecewise function:

\[ f(x) = \begin{cases} x + 1 & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ x - 1 & \text{if } x > 2 \end{cases} \]

We want to find the limit as \( x \to 2 \). As we approach 2 from the left, \( f(x) \) approaches \( 3 \), and as we approach 2 from the right, \( f(x) \) approaches \( 1 \). Since the left-hand limit and right-hand limit are different, the limit does not exist at \( x = 2 \).

Consider the function \( f(x) = \frac{1}{x-1} \). Let’s find the limit as \( x \to 1 \):

\[ \lim_{x \to 1} \frac{1}{x-1} \]

As \( x \) approaches 1 from the right, \( f(x) \to +\infty \), and as \( x \) approaches 1 from the left, \( f(x) \to -\infty \). Because the function approaches infinity (in both directions), the limit does not exist.

Consider the function \( f(x) = \sin\left(\frac{1}{x}\right) \). Let’s find the limit as \( x \to 0 \):

\[ \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \]

As \( x \) approaches 0, the function oscillates between -1 and 1. Since the function does not settle at a single value, the limit does not exist.

1. What is the limit of the function \( f(x) = \frac{2x+1}{x-1} \) as \( x \to 1 \)? Does the limit exist?

2. For the function \( f(x) = \frac{x^2 - 4}{x-2} \), calculate the limit as \( x \to 2 \). Does the limit exist?

3. Why does the limit of \( \sin\left(\frac{1}{x}\right) \) as \( x \to 0 \) not exist? Explain the oscillatory behavior of this function.

In this lesson, we learned about limits and what it means when the limit of a function exists or does not exist. A limit exists when the function approaches a single value as \( x \) approaches a point. If the function behaves erratically, jumps, or approaches infinity, the limit does not exist. Understanding the existence and non-existence of limits is crucial for analyzing the behavior of functions, especially in calculus where limits lead to the concept of derivatives and integrals.