Practice: Continuity & The IVT

Problem 1: Geometric Meaning of the Delta Fence

In the formal definition of continuity, what does the inequality $|x - x_0| < \delta$ represent geometrically on a graph?

A horizontal corridor of width $2\delta$ centered around the y-axis. A vertical corridor (or window) of total width $2\delta$, centered around the target input $x_0$ on the x-axis. A circle of radius $\delta$ around the point $(x_0, f(x_0))$. The slope of the tangent line at $x_0$.

Problem 2: Geometric Meaning of the Epsilon Corridor

Similarly, what does the inequality $|f(x) - f(x_0)| < \epsilon$ represent geometrically?

A horizontal corridor of total height $2\epsilon$, forming a strict ceiling and floor centered around the target output $f(x_0)$. A vertical asymptote at $x = \epsilon$. The maximum rate of change (derivative) allowed. The minimum possible distance between the curve and the x-axis.

Problem 3: The Role of the Adversary

In the $\epsilon$-$\delta$ proof game, what does the adversary's choice of $\epsilon$ represent?

The exact point on the x-axis where we are evaluating the limit. The maximum allowable error boundary around the target output. The steepness of the tangent line. The required precision of our input dial.

Problem 4: The Grand Arc of Calculus

Why did mathematicians spend centuries developing this rigorous $\epsilon$-$\delta$ definition of continuity instead of just trusting the intuition of "drawing a line without lifting your pencil"?

Because drawing lines on paper is imprecise. To ensure that instantaneous rates of change (derivatives) could be calculated safely without accidentally measuring across invisible microscopic holes or jumps. To prove that straight lines are polynomials. Because Isaac Newton demanded it.

Problem 5: Three Pillars of Continuity

A function $f(x)$ is considered fully continuous at a point $x=c$ only if three conditions are met. Which of the following is NOT one of those three conditions?

The function value $f(c)$ must be defined. The overall limit $\lim_{x \to c} f(x)$ must exist. The function must have a derivative (slope) of $0$ at $x=c$. The limit $\lim_{x \to c} f(x)$ must equal the function value $f(c)$.

Problem 6: Translating Quantifiers (Existential)

In the formal definition of continuity ($\forall \epsilon > 0, \exists \delta > 0$), what is the strict logical meaning of the $\exists$ (Existential) quantifier applied to $\delta$?

It means a single, universal $\delta$ value will work for all possible functions. It means the defender only needs to prove that at least one valid, positive $\delta$ can be calculated in response to the specific $\epsilon$ chosen by the adversary. It means the adversary gets to choose both $\epsilon$ and $\delta$. It implies that $\delta$ must always be exactly equal to $\epsilon$.

Problem 7: The Universality of the Formula

In the formal $\epsilon$-$\delta$ definition ($\forall \epsilon > 0, \exists \delta > 0 \dots$), why is the $\forall$ (Universal quantifier) placed on $\epsilon$ instead of $\delta$?

Because the input error ($\delta$) is always larger than the output error ($\epsilon$). Because to prove absolute continuity, the defense must hold up against every conceivable level of precision the adversary demands, no matter how microscopic. Because we are proving the existence of the y-axis. Because the definition was written before the invention of limits.

Problem 8: Shrinking the Delta Fence

You are defending a continuous function. The adversary sets $\epsilon = 0.5$. You calculate that an input tolerance of $\delta = 0.1$ perfectly traps the function. If you decide to be extra cautious and use $\delta = 0.05$ instead, what happens?

The proof breaks because $\delta$ must be exactly equal to the threshold value. The proof still works. Any smaller positive $\delta$ will force the inputs even closer to the target, keeping the outputs safely inside the $\epsilon$-corridor. The output error will double, violating the $\epsilon$ boundary. The function will become discontinuous.

Problem 9: The Epsilon-Delta Machine

You are operating a machine governed by the linear function $f(x) = 5x - 2$. Your perfect target output is $f(3) = 13$. The adversary demands an output tolerance of $\epsilon = 0.5$. What is the required input tolerance $\delta$ around $x_0 = 3$ to mathematically guarantee $|f(x) - 13| < 0.5$?

$\delta = 0.5$ $\delta = 0.1$ $\delta = 2.5$ $\delta = 0.05$

Problem 10: Epsilon-Delta for a Negative Slope

You are testing continuity for $f(x) = -2x + 5$ at $x_0 = 1$. The target output is $f(1) = 3$. If the adversary chooses $\epsilon = 0.4$, what is the strictly required $\delta$?

$\delta = 0.8$ $\delta = -0.2$ $\delta = 0.4$ $\delta = 0.2$

Problem 11: Fractional Linear Delta

For the function $f(x) = \frac{1}{2}x + 4$, the adversary chooses $\epsilon = 0.1$. What $\delta$ is required?

$\delta = 0.1$ $\delta = 0.05$ $\delta = 0.2$ $\delta = 2.0$

Problem 12: Non-Linear Epsilon-Delta Scaling

For a strictly increasing curve that curves upward (like $f(x) = x^2$ for $x > 0$), what happens to the required $\delta$ window as you move further to the right along the x-axis, assuming the adversary's $\epsilon$ remains the same?

The $\delta$ window remains exactly the same size. The $\delta$ window must become wider. The $\delta$ window must become narrower (stricter). $\delta$ becomes negative.

Problem 13: Removable Discontinuities

Suppose a function $g(x)$ is defined such that $\lim_{x \to 3} g(x) = 5$, but the actual evaluated function gives $g(3) = 8$. What type of discontinuity exists at $x = 3$?

An infinite discontinuity (vertical asymptote). A jump discontinuity. A removable discontinuity (a hole with a displaced point). The function is actually continuous.

Problem 14: Jump Discontinuities

If you calculate the left-hand limit of $f(x)$ as $x \to 2^-$ and get $4$, and calculate the right-hand limit as $x \to 2^+$ and get $7$, what can you conclude about the function at $x=2$?

It has a jump discontinuity because the left and right trajectories arrive at different destinations. It has a removable discontinuity (a hole). It is continuous, but not differentiable. The overall limit $\lim_{x\to 2} f(x)$ exists and is $5.5$.

Problem 15: Algebraic Hole Evaluation

Consider $f(x) = \frac{x^2 - 16}{x - 4}$. The function is undefined at $x=4$ (yielding $\frac{0}{0}$). If you factor the numerator and cancel the problematic term, what is the intended limit as $x \to 4$?

$0$ Does not exist (Vertical Asymptote) $8$ $4$

Problem 16: Anatomy of the Discontinuity Jailbreak

Consider the step function $H(x)$ which equals $0$ for $x < 0$ and $1$ for $x \ge 0$. We want to prove it is discontinuous at $x_0 = 0$. The adversary sets $\epsilon = 0.2$ around the target output $H(0) = 1$. Why does the defender's $\delta$-trap fail?

Because $H(0)$ evaluates to $1$, which is larger than $\epsilon$. Because the $\epsilon$-corridor is too wide, allowing the function to escape the top boundary. Because any $\delta$-window around $x=0$, no matter how tiny, will include negative $x$ inputs. These output $0$, which completely misses the $[0.8, 1.2]$ required corridor. Because the domain of the function does not include negative numbers.

Problem 17: Intermediate Value Theorem Guarantees

Suppose $f(x)$ is a strictly continuous function on the closed interval $[1, 5]$. You are given that $f(1) = -4$ and $f(5) = 12$. Which of the following statements is mathematically guaranteed by the IVT?

There is exactly one root where $f(x) = 0$ in the interval $(1, 5)$. There is at least one number $c \in (1, 5)$ such that $f(c) = 7$. The function must be strictly increasing on the interval $[1, 5]$. The function will reach a maximum value of $12$ and a minimum of $-4$.

Problem 18: Real-World IVT

At 8:00 AM, the outside temperature was $15^\circ\text{C}$. At 12:00 PM, the temperature was $25^\circ\text{C}$. Assuming temperature changes continuously, what does the IVT guarantee?

The temperature rose exactly $2.5^\circ\text{C}$ every hour. At some exact moment between 8:00 AM and 12:00 PM, the temperature was exactly $20.3^\circ\text{C}$. The maximum temperature of the day was $25^\circ\text{C}$. The temperature never dropped below $15^\circ\text{C}$ during that time.

Problem 19: The IVT and Polynomials

Why is the Intermediate Value Theorem always safe to use on any standard polynomial function (e.g., $f(x) = x^4 - 3x^2 + 2$) over any closed interval $[a, b]$?

Because polynomials always evaluate to positive numbers. Because polynomials are mathematically continuous everywhere on the real number line ($\mathbb{R}$). Because polynomials are always strictly increasing. Because polynomials never cross the x-axis.

Problem 20: Existence of Roots

Let $f(x) = x^3 + x - 1$. You evaluate $f(0) = -1$ and $f(1) = 1$. What does the Intermediate Value Theorem allow you to conclude?

The equation $x^3 + x - 1 = 0$ has exactly one root on the interval $(0, 1)$. The equation $x^3 + x - 1 = 0$ has at least one root on the interval $(0, 1)$. The limit of the function as $x$ approaches $0.5$ is $0$. The function is discontinuous at $x = 0.5$.

Problem 21: Transcendental Root Proving

Let $f(x) = \cos(x) - x$. You evaluate the interval $[0, \frac{\pi}{2}]$. Given $f(0) = 1$ and $f(\frac{\pi}{2}) = -\frac{\pi}{2}$, what does the IVT conclude?

The IVT cannot be applied because $\cos(x)$ is not algebraic. There is at least one root $c$ in $(0, \frac{\pi}{2})$ where $\cos(c) = c$. The function is undefined at $\frac{\pi}{2}$. The root must be exactly halfway at $\frac{\pi}{4}$.

Problem 22: Exponential IVT

Consider $f(x) = e^x - 4$ on the interval $[0, 2]$. We know $e^0 = 1$ and $e^2 \approx 7.389$. Thus, $f(0) = -3$ and $f(2) \approx 3.389$. Does $e^x = 4$ have a solution in this interval?

Yes, because $f(x)$ is continuous and changes sign from negative to positive. No, because $e$ is an irrational number. No, because exponential functions are strictly increasing. Yes, but only if we expand the interval to include negative numbers.

Problem 23: The IVT Domain Trap

Consider the function $g(x) = \frac{1}{x-3}$. We evaluate the function at $x = 2$ and $x = 4$, finding $g(2) = -1$ and $g(4) = 1$. Because $0$ lies between $-1$ and $1$, does the IVT guarantee that a root $g(c) = 0$ exists in the interval $(2, 4)$?

Yes, because the endpoints have opposite signs. No, because rational functions cannot have roots. No, because $g(x)$ is undefined and discontinuous at $x = 3$, which lies inside the interval $[2, 4]$, shattering the IVT requirement. Yes, because the function is continuous at both $x=2$ and $x=4$.

Problem 24: Same-Sign IVT Trap

You test a continuous function $f(x)$ on $[2, 6]$ and find $f(2) = 4$ and $f(6) = 10$. Both outputs are positive. Does this guarantee that $f(x)$ has NO roots in the interval?

Yes, because the IVT proves the function stays above the x-axis. No, the IVT simply provides no guarantee either way. The curve could dip below zero and come back up, or it could stay entirely positive. Yes, because continuous functions cannot change direction. No, because the IVT always guarantees at least one root.

Problem 25: Bisection Method Logic

You are using the bisection algorithm to find a root of a continuous function $f(x)$ on the interval $[0, 4]$. You know $f(0)$ is negative and $f(4)$ is positive. You check the midpoint $c = 2$, and find that $f(2)$ is positive. What is your new, smaller interval guaranteed to contain the root?

$[2, 4]$ $[0, 4]$ $[0, 2]$ The root no longer exists.

Practice Problems: Continuity & The IVT