Foundations of Limits and Continuity

TL;DR

Limits describe what a function's output approaches as its input gets closer to a certain value, without necessarily reaching it. Continuity means a function's graph has no breaks, jumps, or holes, essentially being drawable without lifting your pencil. Understanding limits is fundamental to grasping continuity and later, calculus concepts like derivatives and integrals.

1. The Mental Model

Think of a limit like approaching a destination: you're getting closer and closer, but you might not actually step on the exact spot. Continuity is like a smooth, unbroken road—you can drive from one point to another without any sudden drops or barriers.

2. The Core Material

What's a Limit?

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A limit tells you the value a function tends towards as its input variable gets arbitrarily close to a particular number. It's not necessarily the function's actual value at that number. We write this as $\lim_{x \to c} f(x) = L$, meaning "the limit of $f(x)$ as $x$ approaches $c$ is $L$."

The key idea is "approaches." You're checking the behavior of the function on either side of $c$. For a limit to exist, the function must approach the same value from both the left and the right.

One-Sided Limits

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• Left-Hand Limit: $\lim_{x \to c^-} f(x) = L_1$ means as $x$ approaches $c$ from values less than $c$.
• Right-Hand Limit: $\lim_{x \to c^+} f(x) = L_2$ means as $x$ approaches $c$ from values greater than $c$.

A two-sided limit $\lim_{x \to c} f(x)$ exists if and only if $L_1 = L_2$. If they're different, the limit doesn't exist.

Properties of Limits

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Limits behave nicely with arithmetic operations:
- $\lim_{x \to c} (f(x) \pm g(x)) = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$
- $\lim_{x \to c} (f(x) \cdot g(x)) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)$
- $\lim_{x \to c} (f(x) / g(x)) = \lim_{x \to c} f(x) / \lim_{x \to c} g(x)$, provided $\lim_{x \to c} g(x) \neq 0$
- $\lim_{x \to c} k \cdot f(x) = k \cdot \lim_{x \to c} f(x)$ (where $k$ is a constant)

For most "nice" functions (like polynomials, exponentials, sines, cosines), you can often find the limit by simply plugging the value $c$ into the function. This is called direct substitution. Problems arise when direct substitution leads to indeterminate forms like $0/0$ or $k/0$.

What's Continuity?

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A function $f(x)$ is continuous at a point $x=c$ if three conditions are met:
1. $f(c)$ exists: The function is defined at $c$.
2. $\lim_{x \to c} f(x)$ exists: The limit as $x$ approaches $c$ exists.
3. $\lim_{x \to c} f(x) = f(c)$: The limit value is equal to the function's actual value at $c$.

If any of these conditions fail, the function is discontinuous at $c$. If a function is continuous at every point in its domain, we say it's continuous everywhere.

Here's how these ideas relate:

graph TD
    A["Function $f(x)$ is defined at $x=c$ ($f(c)$ exists)?"] -->|No| B["NOT Continuous at $x=c$"]
    A -->|Yes| C["Does $\lim_{x \to c} f(x)$ exist?"]
    C -->|No| B
    C -->|Yes| D["Is $\lim_{x \to c} f(x) = f(c)$?"]
    D -->|No| B
    D -->|Yes| E["Continuous at $x=c$"]

Types of Discontinuities

• Removable (Hole): Condition 3 fails, usually because $f(c)$ is undefined or differs from the limit, but the limit exists. Example: $f(x) = (x^2 - 1) / (x - 1)$ at $x=1$. You can "remove" it by redefining the function at that point.
• Non-Removable (Jump): Condition 2 fails because the left-hand limit and right-hand limit are different. Example: Piecewise functions where the two pieces don't meet.
• Non-Removable (Infinite): Condition 2 or 1 fails because the function approaches $\pm \infty$ at $c$. Example: Vertical asymptotes like $f(x) = 1/x$ at $x=0$.

3. Worked Example

Let's examine the function $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ 3x - 1 & \text{if } x > 2 \end{cases}$ for continuity at $x=2$.

We need to check the three conditions:

1. Does $f(2)$ exist?
   From the definition, when $x=2$, $f(2) = 5$. So, yes, it exists.

2. Does $\lim_{x \to 2} f(x)$ exist?
   We need to check the left and right-hand limits:

   • Left-hand limit: As $x \to 2^-$, we use the $x^2 + 1$ part of the function:
     $\lim_{x \to 2^-} (x^2 + 1) = (2)^2 + 1 = 4 + 1 = 5$.
   • Right-hand limit: As $x \to 2^+$, we use the $3x - 1$ part of the function:
     $\lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 6 - 1 = 5$.
     Since the left-hand limit (5) equals the right-hand limit (5), $\lim_{x \to 2} f(x) = 5$ exists.

3. Is $\lim_{x \to 2} f(x) = f(2)$?
   We found $\lim_{x \to 2} f(x) = 5$ and $f(2) = 5$. Since $5 = 5$, this condition is met.

All three conditions are satisfied, so $f(x)$ is continuous at $x=2$.

4. Key Takeaways

• A limit describes a function's intended value, not necessarily its actual value, as the input approaches a specific point.
• For a limit to exist at a point, the function must approach the same value from both the left and the right sides.
• Direct substitution is a common way to find limits for "well-behaved" functions where the denominator won't be zero.
• Continuity at a point means the function is defined there, the limit exists there, and these two values are equal.
• Visualizing a continuous function means you can draw its graph without lifting your pencil.
• Discontinuities can be removable (a hole) or non-removable (a jump or an infinite break).

Common Mistakes to Avoid

• Confusing $f(c)$ with $\lim_{x \to c} f(x)$; they aren't always the same.
• Assuming a limit exists just because the function is defined at that point.
• Forgetting to check both one-sided limits when direct substitution fails or for piecewise functions.
• Not checking all three conditions for continuity; missing one means you can't conclude continuity.

5. Now Try It

Consider the function $g(x) = \frac{x^2 - 4}{x - 2}$ and the point $x=2$.
1. Find $g(2)$. What happens?
2. Find $\lim_{x \to 2} g(x)$ by simplifying the function first.
3. Based on your findings, describe the type of discontinuity (if any) at $x=2$ and explain why it's that type.

Success looks like: You correctly identify that $g(2)$ is undefined, but you're still able to find a limit value. You then correctly classify the discontinuity as removable and explain how you know.