Removable: can be 'fixed' by redefining the function. Non-removable: cannot be fixed; often vertical asymptotes.

Both indicate the function approaches infinity, but the first approaches from the right, the second from the left.

The first approaches positive infinity from the right, the second approaches negative infinity from the left, indicating a vertical asymptote.

A limit exists if the function approaches a finite value. A vertical asymptote exists if the function approaches infinity.

Near a vertical asymptote, the function approaches infinity. Near a hole, the function approaches a finite value, but is undefined at that point.

f(x) has different signs on either side of x=0. g(x) is always positive.

f(x) is defined for x > 0. g(x) is defined for all x except x = 0.

Polynomial functions do not have vertical asymptotes, while rational functions can have vertical asymptotes where the denominator is zero.

The limit of f(x) is infinite, indicating a vertical asymptote. The limit of g(x) is 0.

Even power: function approaches same infinity from both sides. Odd power: function approaches opposite infinities from each side.

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 $x=2$.

It suggests that $f'(x)$ may also have a vertical asymptote or be unbounded near that x-value.

Look for a vertical line that the function approaches but never crosses; the function's value tends towards $\pm \infty$ near this line.

It shows that $\lim_{x\to 0^+} f(x) = \infty$ and $\lim_{x\to 0^-} f(x) = -\infty$, indicating a vertical asymptote at $x=0$.

The graph approaches the y-axis ($x=0$) very closely as $x$ approaches 0 from the right, and the function's value goes to negative infinity.

If the graph goes up towards the asymptote, the limit is positive infinity; if it goes down, the limit is negative infinity.

The function is not defined at x=a, so x=a is not in the domain.

The steeper the graph gets as it approaches the asymptote, the faster the function is approaching infinity (either positive or negative).

A hole in the graph, indicating a point where the function is undefined but could be defined to make the function continuous.

Both limits indicate the graph goes up near x=a. The first approaches from the right, the second from the left.

The graph will approach the vertical line x=a very closely, with the y-values either increasing or decreasing without bound.