Defining Continuity at a Point - AP Calculus BC

A function is continuous at an endpoint if $\text{lim}$ from the interior equals the endpoint value. Only the one-sided limit from inside the interval needs to match.

Yes, $f(x) = \text{ln}(x)$ is continuous for $x > 0$. Logarithm is defined and smooth for all positive real numbers.

A function without any discontinuities over its entire domain. No breaks, jumps, or holes exist anywhere in the domain.

The limits from each piece must equal the function's value at the boundary. Left and right limits at transition points must equal the function value.

$\text{lim}_{x \to a^+} f(x) = f(a)$. The right-hand limit must equal the function value at the point.

Yes, $f(x) = |x|$ is continuous at $x = 0$. Both one-sided limits equal 0, which equals $f(0) = |0| = 0$.

No, $f(x) = x^2$ is continuous everywhere. Polynomial functions are continuous at every point in their domain.

$\text{lim}_{x \to a} f(x)$ does not exist. When the limit fails to exist, the discontinuity cannot be removed.

The limit is $2$, removable discontinuity at $x = 1$. Factor $(x-1)(x+1)$ and cancel to get $\lim_{x \to 1} (x+1) = 2$.

No, removable discontinuity at $x = 1$. Function is undefined at $x=1$ where the denominator equals zero.

$\text{lim}{x \to 2} \frac{x^2 - 4}{x - 2} = 4$. Removable discontinuity at $x = 2$. Factor and cancel to get $\lim{x \to 2} (x+2) = 4$; undefined at $x=2$.

Non-removable discontinuity. The discontinuity cannot be fixed by redefining a single point.

A discontinuity where $\text{lim}{x \to a} f(x)$ exists but $\text{lim}{x \to a} f(x) \neq f(a)$. The gap can be filled by redefining the function at that point.

A discontinuity where $\text{lim}{x \to a^+} f(x) \neq \text{lim}{x \to a^-} f(x)$. The function has different left and right limit values at the point.

Yes, $f(x) = e^x$ is continuous everywhere. Exponential functions are continuous throughout their entire domain.

$\text{lim}_{x \to 0} \frac{\text{sin}x}{x} = 1$. Continuous at $x = 0$. This is a standard limit; $\frac{\sin x}{x}$ approaches 1 as $x \to 0$.

A function $f(x)$ is continuous at $x = a$ if $\text{lim}_{x \to a} f(x) = f(a)$. This is the formal definition combining limit existence and function value equality.

Both $\text{lim}{x \to a^+} f(x)$ and $\text{lim}{x \to a^-} f(x)$ must exist. Each directional approach must have a finite limit value.

Yes, $f(x) = 2x + 3$ is continuous for all $x$. Linear functions are continuous everywhere in the real numbers.

$\text{lim}{x \to a^+} f(x) = \text{lim}{x \to a^-} f(x) = f(a)$. Both one-sided limits must exist and equal the function value.

Removable discontinuity at $x = 3$. The limit exists but the function can be redefined to remove the gap.

A discontinuity where $\text{lim}_{x \to a} f(x)$ does not exist. The limit failure creates an unfixable discontinuity.

The limit does not exist. Non-removable discontinuity at $x = 0$. The function approaches infinity, so the limit doesn't exist.

No, infinite discontinuity at $x = 0$. Function approaches positive infinity at $x=0$, creating infinite discontinuity.

A function is continuous on an interval if it is continuous at every point in the interval. Every point in the interval must satisfy the continuity definition.

The limit must exist and be equal to $f(a)$. The limit and function value must be identical for continuity.

A discontinuity where $f(x)$ approaches $\text{infinity}$ as $x$ approaches $a$. The function grows without bound as it approaches the point.

$\text{lim}_{x \to a^-} f(x) = f(a)$. The left-hand limit must equal the function value at the point.

Non-removable (infinite) discontinuity at $x = 0$. Function approaches infinity at $x=0$, creating an unbounded discontinuity.