Introduction to Limits and Continuity

TL;DR

Limits describe what a function's output gets closer to as its input approaches a certain value, even if the function isn't defined at that exact point. Continuity means you can draw a function's graph without lifting your pencil, which mathematically depends on the limit matching the function's actual value. Understanding these concepts is fundamental for calculus, especially for derivatives.

1. The Mental Model

Think of a limit as predicting where you'll land if you keep walking towards a spot, even if there's a tiny hole exactly at that spot. Continuity just means there are no such holes, jumps, or breaks in your path.

2. The Core Material

What is a Limit?

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A limit tells us what value a function approaches as its input gets closer and closer to some number. It doesn't care what the function actually is at that exact number, only what it's tending towards.

Let's say we have a function $f(x)$. We write the limit as:
$\lim_{x \to c} f(x) = L$

This means "as $x$ gets really, really close to $c$ (from both sides), $f(x)$ gets really, really close to $L$."

Why do we need limits?

Sometimes a function might be undefined at a certain point (like dividing by zero). Limits help us understand the function's behavior around that point. For example, if $f(x) = \frac{x^2 - 1}{x - 1}$, you can't plug in $x=1$ because you'd get $\frac{0}{0}$. But what happens as $x$ gets super close to 1?

As you can see, as $x$ gets closer to 1, $f(x)$ gets closer to 2. So, $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$.

Left-Hand and Right-Hand Limits

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For a limit to exist, the function must approach the same value whether you're coming from the left (numbers smaller than $c$) or the right (numbers larger than $c$).
* Left-hand limit: $\lim_{x \to c^-} f(x)$
* Right-hand limit: $\lim_{x \to c^+} f(x)$

If $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L$, then $\lim_{x \to c} f(x) = L$. If they're different, the overall limit doesn't exist.

What is 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 that point. There's no hole.
2. $\lim_{x \to c} f(x)$ exists: The limit as $x$ approaches $c$ exists (meaning left and right limits match). There's no jump or break where the two sides don't meet.
3. $\lim_{x \to c} f(x) = f(c)$: The limit value is exactly equal to the function's actual value at that point. The path leads exactly to where the function is defined.

If any of these conditions fail, the function is discontinuous at $c$.

graph TD
    A["Function $f(x)$ is continuous at $x=c$?"] --> B{"Is $f(c)$ defined?"}
    B -- Yes --> C{"Does $\lim_{x \to c} f(x)$ exist?"}
    C -- Yes --> D{"Is $\lim_{x \to c} f(x) = f(c)$?"}
    D -- Yes --> E["Function is Continuous at $x=c$"]
    B -- No --> F["Function is Discontinuous (Hole / Undefined)"]
    C -- No --> G["Function is Discontinuous (Jump / Break)"]
    D -- No --> H["Function is Discontinuous (Removable Discontinuity / Hole)"]

Types of Discontinuity

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• Removable (Hole): The limit exists, but $f(c)$ is either undefined or doesn't match the limit. Imagine a graph with a single point missing, or a single point misplaced.
• Jump: The left-hand limit and right-hand limit are different. The graph "jumps" from one value to another.
• Infinite: The function shoots off to positive or negative infinity as $x$ approaches $c$ (e.g., vertical asymptotes).

3. Worked Example

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

We need to check the three conditions for continuity:

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

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

   • Left-hand limit: As $x$ approaches 2 from the left ($x<2$), we use $g(x) = x+1$.
     $\lim_{x \to 2^-} (x+1) = 2+1 = 3$.
   • Right-hand limit: As $x$ approaches 2 from the right ($x>2$), we use $g(x) = x^2-1$.
     $\lim_{x \to 2^+} (x^2-1) = 2^2-1 = 4-1 = 3$.
     Since the left-hand limit (3) equals the right-hand limit (3), $\lim_{x \to 2} g(x) = 3$. Yes, it exists.

3. Is $\lim_{x \to 2} g(x) = g(2)$?
   We found $\lim_{x \to 2} g(x) = 3$ and $g(2) = 5$.
   Since $3 \neq 5$, this condition is not met.

Therefore, the function $g(x)$ is discontinuous at $x=2$. It has a removable discontinuity because the limit exists, but the function's value at that point is different.

4. Key Takeaways

• A limit describes the value a function approaches, not necessarily its actual value at that point.
• For a limit to exist at a point, the function must approach the same value from both the left and the right sides.
• Continuity at a point means the function is defined there, the limit exists there, and these two values are equal.
• Visually, a continuous function can be drawn without lifting your pencil from the paper.
• Discontinuities can be holes (removable), jumps, or infinite (vertical asymptotes).

Common Mistakes to Avoid:
- Don't assume a limit exists just because the function is defined at that point.
- Forgetting to check both left-hand and right-hand limits when evaluating if a limit exists, especially for piecewise functions.
- Confusing "the limit exists" with "the function is continuous" – continuity requires an extra step.
- Incorrectly simplifying expressions before evaluating limits, potentially overlooking critical points where the original expression was undefined.

5. Now Try It

Consider the function $h(x) = \frac{x^2 - 4}{x - 2}$.
1. Find $\lim_{x \to 2} h(x)$.
2. Determine if $h(x)$ is continuous at $x=2$. If not, explain why.

What success looks like: You should find the limit by simplifying the expression before plugging in the value. You'll then clearly state whether the function is continuous at $x=2$ and reference the specific condition(s) for continuity that are met or failed.