# Continuity

Mathematics I · Differential Calculus · https://tryals.app/learn/mathematics-i/continuity

## Defining Continuity

A function is **continuous** at a point $a$ when you can draw it through $a$ without lifting your pen. Formally, three conditions must all hold simultaneously:

| Condition | Requirement |
| --- | --- |
| 1. Value | $f(a)$ is **defined** |
| 2. Limit | $\lim_{x \to a} f(x)$ **exists** |
| 3. Agreement | $\lim_{x \to a} f(x) = f(a)$ |

Polynomials, $\sin$, $\cos$, and exponentials are continuous everywhere, as are sums, products, and compositions of continuous functions.

**Common pitfall:** Thinking a function is continuous merely because its **limit exists** at a point. A **removable discontinuity** (a hole) has a perfectly good limit yet is **not** continuous, because $f(a)$ is missing or mismatched. Existence of the limit is necessary but not sufficient.

## Types and Theorems

When continuity fails, the **discontinuity** falls into three classic types:

| Type | Description |
| --- | --- |
| **Removable** | A single hole; limit exists but $f(a)$ is missing or misplaced. |
| **Jump** | One-sided limits exist but differ; the graph leaps. |
| **Infinite** | The function blows up to $\pm\infty$ at a vertical asymptote. |

Continuity powers two cornerstone theorems on a closed interval $[a, b]$:

- The **Intermediate Value Theorem**: Takes every value between $f(a)$ and $f(b)$. If $f$ changes sign, it has a **root** in between (basis of bisection).
- The **Extreme Value Theorem**: Attains a **maximum** and **minimum**, guaranteeing optimization solutions exist.

These theorems fail entirely without continuity.

## Practice questions

5 of this lesson's 10 practice questions, with answers. The full set is in the app.

### 1. Continuity at a point requires three separate conditions, and the limit existing is only one of them.

**Answer:** True

**Why:** True. $f(a)$ must be defined, $\lim_{x\to a} f(x)$ must exist, and the two must be equal. A removable discontinuity satisfies the second and fails the third; a function undefined at $a$ fails the first. Each condition rules out a different failure, which is why the definition lists all three.

Page: https://tryals.app/practice/mathematics-i/continuity/continuity-at-a-point-requires-three-separate-conditions-and-the

### 2. Sort each function by whether it is continuous at every real number.

**Answer:**

- Continuous everywhere: x cubed - 2x + 1, cos x, e to the power x, the absolute value of x
- Not continuous everywhere: 1/(x - 2), (x + 1)/(x squared - 9)

**Why:** The two quotients break where their denominators vanish, at x = 2 and at x = 3 and -3. The absolute value is the interesting one: it has a sharp corner at 0 yet is perfectly continuous there, the corner will block DIFFERENTIABILITY in the next lesson, which is a strictly stronger demand.

Page: https://tryals.app/practice/mathematics-i/continuity/sort-each-function-by-whether-it-is-continuous-at-every-real-number

### 3. Order the checks that establish continuity of f at the point a.

**Answer:**

1. Check that f(a) is defined
2. Check that the limit of f(x) as x approaches a exists
3. Check that the limit equals f(a)
4. Declare f continuous at a

**Why:** The order is forced: comparing a value with a limit is meaningless unless both exist. A removable discontinuity passes the second check and fails the first or third, which is why the lesson insists all three conditions be verified rather than just the limit.

Page: https://tryals.app/practice/mathematics-i/continuity/order-the-checks-that-establish-continuity-of-f-at-the-point-a

### 4. The function |x| is continuous everywhere and has no derivative at 0. What does that pair of facts establish about the two ideas?

A. That continuity is strictly weaker than differentiability
B. That continuity and differentiability are unrelated properties
C. That |x| is not genuinely continuous at 0, only appears so
D. That the derivative exists but equals zero at the corner

**Answer:** A. That continuity is strictly weaker than differentiability

**Why:** The two one-sided slopes are −1 and +1 and never agree, so no tangent exists at a point where the curve is perfectly unbroken. Weierstrass pushed this to its limit with a function continuous everywhere and differentiable nowhere, all corners, which is far stranger than the picture suggests.

Page: https://tryals.app/practice/mathematics-i/continuity/the-function-x-is-continuous-everywhere-and-has-no-derivative-at-0

### 5. f is continuous on the closed interval from 0 to 4. Select every statement that MUST be true.

A. f attains a minimum value on that interval
B. f takes every value between f(0) and f(4)
C. f attains a maximum value on that interval
D. f has a root somewhere in the interval
E. f is differentiable on the interval

**Answer:** A. f attains a minimum value on that interval; B. f takes every value between f(0) and f(4); C. f attains a maximum value on that interval

**Why:** The Extreme Value Theorem supplies a maximum and a minimum, and the Intermediate Value Theorem supplies every intermediate value. But a root needs a SIGN CHANGE, which was not assumed, the constant function 7 has none, and continuity never implies differentiability, as the absolute value shows.

Page: https://tryals.app/practice/mathematics-i/continuity/f-is-continuous-on-the-closed-interval-from-0-to-4-select-every