A vertical line $x=a$ that a function approaches but never touches; function is unbounded at $x=a$.

A point where a function is undefined or behaves erratically.

Limits that evaluate to either positive or negative infinity.

The function's value grows or shrinks without any bound, approaching infinity or negative infinity.

A vertical asymptote is a type of discontinuity where the function is undefined.

As $x$ approaches $a$, the value of $f(x)$ increases without bound.

As $x$ approaches $a$, the value of $f(x)$ decreases without bound.

A limit that considers the function's behavior as x approaches a value from either the left or the right.

The limit of $f(x)$ as $x$ approaches $a$ from the right.

The limit of $f(x)$ as $x$ approaches $a$ from the left.

$\lim_{x\to a} f(x) = \pm\infty$, $\lim_{x\to a^+} f(x) = \pm\infty$, or $\lim_{x\to a^-} f(x) = \pm\infty$

If $\lim_{x\to a^+} f(x) = \pm \infty$ or $\lim_{x\to a^-} f(x) = \pm \infty$, then $x=a$ is a vertical asymptote.

$\lim_{x\to 0^+} ln(x) = -\infty$

f(x) = $\frac{g(x)}{x-a}$, where g(a) != 0

x = -c

$\infty$

$\lim_{x \to a^+} f(x) = \infty$ and $\lim_{x \to a^-} f(x) = -\infty$

$\lim_{x\to 0^+} \frac{1}{x} = \infty$

$\lim_{x\to 0^-} \frac{1}{x} = -\infty$

$q(a) = 0$ and $p(a) \neq 0$

1. Factor numerator and denominator. 2. Simplify the function. 3. Find values where the denominator is zero and the numerator is non-zero.

1. Evaluate $\lim_{x\to a^+} f(x)$. 2. Evaluate $\lim_{x\to a^-} f(x)$. 3. Show that at least one of these limits is $\pm \infty$.

1. Factor and simplify the function. 2. If a factor cancels, it's a removable discontinuity. 3. If a factor remains in the denominator, it's a vertical asymptote.

1. Factor: $f(x) = \frac{x+2}{(x+2)(x-2)}$. 2. Simplify: $f(x) = \frac{1}{x-2}$. 3. VA at $x=2$.

1. Rewrite as $f(x) = \frac{\sin(x)}{\cos(x)}$. 2. Find where $\cos(x) = 0$. 3. $x = \frac{(2n+1)\pi}{2}$, where n is an integer.

1. Factor: $f(x) = \frac{(x-1)(x+1)}{x-1}$. 2. Simplify: $f(x) = x+1$. 3. Removable discontinuity at x=1, no vertical asymptote.

1. Determine the sign of the function as x approaches from the left and right. 2. The limit will be either positive or negative infinity.

1. Set denominator equal to zero: $x^2 + 1 = 0$. 2. Solve for x: $x^2 = -1$. 3. No real solutions, so no vertical asymptotes.

1. Check where the denominator is zero: $x=0$. 2. Verify the numerator is not zero at $x=0$: $e^0 = 1 \neq 0$. 3. Vertical asymptote at $x=0$.

1. Check limit from the right: $\lim_{x\to 2^+} \frac{1}{x-2} = \infty$. 2. Check limit from the left: $\lim_{x\to 2^-} \frac{1}{x-2} = -\infty$. 3. Vertical asymptote at $x=2$.