What is a removable discontinuity?

A removable discontinuity is a 'hole' in the graph. It happens when the limit exists (both sides meet), but the actual function value is undefined or located somewhere else. It's called 'removable' because you could easily fix it by redefining just that single point.

Do I have to check continuity everywhere?

Usually, no. Most standard functions (like polynomials, exponentials, and sines/cosines) are continuous everywhere in their domain. You typically only need to manually check 'suspicious' points, like the breaks in piecewise functions, or denominators that might equal zero.

Can a function be continuous but not differentiable?

Yes! Continuity is required for a function to be differentiable (meaning it has a derivative), but it's not enough on its own. A sharp corner on a graph, like at $x = 0$ for the absolute value function $f(x) = |x|$, is continuous (no breaks) but not differentiable (you can't draw a single tangent line there).

What does continuity mean and how do you check it?

In simple terms, a function is continuous if you can draw its entire graph without ever picking up your pencil. If there are no holes, gaps, or sudden jumps, the function is continuous.

In calculus, we can't just rely on looking at a graph to prove this, so we use limits. Checking continuity means verifying that as you get closer to a specific $x$ -value from both sides, the function points exactly to the value that actually exists at that spot.

The Everyday Analogy

Think of continuity like building a bridge across a river. The left side of the bridge is the limit approaching from the left, and the right side is the limit approaching from the right.

For the bridge to be continuous (safe to cross), three things must happen. First, there must actually be a meeting point (the function value exists). Second, both sides of the bridge must aim for the exact same spot (the limit exists). Finally, the sides must actually connect right at that meeting point, with no step up or down.

The Three-Part Checklist

To officially prove a function $f(x)$ is continuous at a point $x = c$ , you must check three specific conditions. If a function fails even one of these, it is not continuous at that point.

Condition 1: $f(c)$ must be defined. This means there is an actual point on the graph at $x = c$ , not a hole or a vertical asymptote.

Condition 2: $\lim_{x \to c} f(x)$ must exist. This means the left-hand limit and the right-hand limit approach the same exact $y$ -value.

Condition 3: $\lim_{x \to c} f(x) = f(c)$ . The limit (where the function is heading) must perfectly match the function's actual value (where the point actually is).

Where Students Slip Up

The most common mistake is stopping after Condition 2. A student will find that the limit exists and assume the function is continuous. But a function can have a limit and still have a "hole" if the actual point is moved somewhere else or missing entirely.

Another trap is piecewise functions. When checking continuity at the "break" between two pieces, you absolutely must calculate the left-hand limit and right-hand limit separately to ensure they meet. Don't just plug the number into one piece and call it a day.

Worked through

Determine if the following piecewise function is continuous at $x = 2$ :

$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ 3x - 1 & \text{if } x > 2 \end{cases}$

We will use the three-part checklist at $c = 2$ .

Step 1: Is $f(2)$ defined? Looking at the middle condition of our piecewise function, when $x = 2$ , $f(x) = 5$ . So, $f(2) = 5$ . Condition 1 is met.

Step 2: Does the limit as $x \to 2$ exist? We need to check the left and right limits. Left-hand limit ( $x < 2$ ): $\lim_{x \to 2^-} (x^2 + 1) = (2)^2 + 1 = 5$ . Right-hand limit ( $x > 2$ ): $\lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5$ . Since the left limit equals the right limit, $\lim_{x \to 2} f(x) = 5$ . Condition 2 is met.

Step 3: Does the limit equal the function value? We found that $\lim_{x \to 2} f(x) = 5$ and $f(2) = 5$ . Since $5 = 5$ , Condition 3 is met.

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

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Where this comes from: OpenStax Calculus Volume 1, Chapter 2: Limits · Stewart Calculus, Chapter 2: Limits and Derivatives · Khan Academy, AP Calculus AB: Continuity