Limits, Continuity and the Intermediate Value Theorem

TL;DR

Limits describe what a function approaches, even if it doesn't actually reach that value, and they're fundamental to calculus. Continuity means a function's graph has no breaks, jumps, or holes. The Intermediate Value Theorem (IVT) uses continuity to guarantee a function hits every value between any two points.

1. The Mental Model

Imagine you're driving towards a destination. The limit is where you're headed, even if you never quite get there (maybe there's a roadblock right at the end). Continuity means your road is smooth, with no surprise gaps or sudden drops. The Intermediate Value Theorem basically says if your road is continuous, and you start at one elevation and end at another, you must have passed through all the elevations in between.

2. The Core Material

What is a Limit?

A limit tells us the value that a function "approaches" as the input (x-value) gets closer and closer to some number. It doesn't care what the function actually does at that exact x-value, only what it's headed towards.

We write it like this: $\lim_{x \to c} f(x) = L$

This reads as "the limit of $f(x)$ as $x$ approaches $c$ is $L$."

Think of it like this:
* From the left: $x$ gets closer to $c$ from values less than $c$. We write: $\lim_{x \to c^-} f(x)$
* From the right: $x$ gets closer to $c$ from values greater than $c$. We write: $\lim_{x \to c^+} f(x)$
* For the overall limit to exist ($\lim_{x \to c} f(x)$), the limit from the left must equal the limit from the right. If they don't match, the limit does not exist (DNE).

What is Continuity?

A function $f(x)$ is continuous at a point $x=c$ if its graph has no breaks, jumps, or holes at that point. If you can draw it without lifting your pencil, it's continuous!

Mathematically, a function $f(x)$ is continuous at $x=c$ if all three of these conditions are met:
1. $f(c)$ is defined (there's a point there).
2. $\lim_{x \to c} f(x)$ exists (the limit approaches a specific value).
3. $\lim_{x \to c} f(x) = f(c)$ (the limit is the point).

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

Here's how to think about different types of discontinuities:

graph TD
    A["Function Discontinuous at x=c"] --> B["Jump Discontinuity"]
    A --> C["Removable Discontinuity (Hole)"]
    A --> D["Infinite Discontinuity (Vertical Asymptote)"]

    B -- "Left and Right Limits Exist" --> B1["But are not equal"]
    C -- "Limit Exists" --> C1["But f(c) is undefined OR limit != f(c)"]
    D -- "Limit is +/- infinity" --> D1["Typically from division by zero"]

The Intermediate Value Theorem (IVT)

The IVT is a cool existence theorem. It doesn't tell you what the value is, or where it happens, but it guarantees that it does happen.

Conditions for IVT:
1. The function $f(x)$ must be continuous on a closed interval $[a, b]$.
2. Let $N$ be any number between $f(a)$ and $f(b)$ (meaning $f(a) < N < f(b)$ or $f(b) < N < f(a)$).

Conclusion of IVT:
Then there must exist at least one number $c$ in the open interval $(a, b)$ such that $f(c) = N$.

Put simply: If you have a continuous function over an interval, it must take on every single y-value between its starting and ending y-values.

3. Worked Example

Let's use a function and check for continuity and apply the IVT.

Consider the function $f(x) = x^2 - 4x + 3$.

Part 1: Is $f(x)$ continuous at $x=2$?

1. Is $f(2)$ defined?
   $f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$. Yes, it's defined.

2. Does $\lim_{x \to 2} f(x)$ exist?
   Since $f(x)$ is a polynomial, it's continuous everywhere, so the limit will exist and be equal to the function's value.
   $\lim_{x \to 2^+} (x^2 - 4x + 3) = (2)^2 - 4(2) + 3 = -1$
   $\lim_{x \to 2^-} (x^2 - 4x + 3) = (2)^2 - 4(2) + 3 = -1$
   Since the left and right limits are equal, $\lim_{x \to 2} f(x) = -1$. Yes, it exists.

3. Does $\lim_{x \to 2} f(x) = f(2)$?
   Yes, $-1 = -1$.

Since all three conditions are met, $f(x)$ is continuous at $x=2$. (In fact, all polynomials are continuous everywhere).

Part 2: Apply the IVT to $f(x) = x^2 - 4x + 3$ on the interval $[0, 5]$. Does $f(x)$ ever equal 5 within this interval?

1. Is $f(x)$ continuous on $[0, 5]$?
   Yes, $f(x)$ is a polynomial, so it's continuous everywhere, including $[0, 5]$.

2. Find $f(a)$ and $f(b)$:
   $f(0) = (0)^2 - 4(0) + 3 = 3$
   $f(5) = (5)^2 - 4(5) + 3 = 25 - 20 + 3 = 8$

3. Check if $N=5$ is between $f(0)$ and $f(5)$:
   We have $f(0) = 3$ and $f(5) = 8$.
   Is $3 < 5 < 8$? Yes.

Conclusion:
Since $f(x)$ is continuous on $[0, 5]$ and $N = 5$ is between $f(0)=3$ and $f(5)=8$, the IVT guarantees that there exists at least one $c$ in $(0, 5)$ such that $f(c) = 5$. We don't know what $c$ is, just that it exists.

(If you were curious, $x^2 - 4x + 3 = 5 \implies x^2 - 4x - 2 = 0$. Using the quadratic formula, $x = \frac{4 \pm \sqrt{16 - 4(1)(-2)}}{2} = \frac{4 \pm \sqrt{24}}{2} = 2 \pm \sqrt{6}$. $2 + \sqrt{6} \approx 4.45$ which is in $(0,5)$.)

4. Key Takeaways

• A limit is what a function approaches, not necessarily what it is, at a specific point.
• For a limit to exist, the left-hand limit must equal the right-hand limit.
• Continuity means a function has no breaks; it's defined, its limit exists, and the limit equals the function's value at that point.
• Polynomials, exponential functions, sine, and cosine are continuous everywhere. Rational functions are continuous on their domains (where the denominator isn't zero).
• The Intermediate Value Theorem requires a continuous function on a closed interval to guarantee it hits all y-values between the endpoints.

Common Mistakes to Avoid:
- Assuming continuity: Always check the three conditions for continuity. Don't just assume a function is continuous, especially at problem points like zeroes in the denominator.
- Confusing limit and function value: $\lim_{x \to c} f(x)$ doesn't have to equal $f(c)$ unless the function is continuous there.
- Ignoring IVT conditions: You must state that the function is continuous on the interval, and that $N$ is between $f(a)$ and $f(b)$ to use the IVT. Without continuity, the theorem doesn't apply.
- "The limit is undefined": Limits don't become "undefined." They either exist (to a number or $\pm\infty$), or they "do not exist" (DNE), often because left and right limits are different.

5. Now Try It

Consider the function $g(x) = \frac{x^2 - 9}{x - 3}$.
1. Determine if $g(x)$ is continuous at $x=3$. If not, explain which condition(s) for continuity fail.
2. Does the $\lim_{x \to 3} g(x)$ exist? If so, what is it?
3. Would the IVT apply to $g(x)$ on the interval $[2, 4]$ to determine if $g(x)=6$ somewhere on that interval? Explain why or why not.

Success looks like correctly identifying the discontinuity, finding the limit despite the discontinuity, and explaining why the IVT cannot be directly applied due to the lack of continuity on the specified interval.