What is a Limit?

`color(black)(lim_(xtoc)f(x))` is what the function, f(x) would equal or be as x approaches c. Even if f(x) is not defined at x=c, the `color(black)(lim_(xtoc)f(x))` would be what f(x) would have been if f(c) was defined.

`color(black)(lim_(xtoc)f(x))` only exists if `color(black)(lim_(xtoc-)f(x)=lim_(xtoc+)f(x))`. `color(black)(lim_(xtoc-)f(x))` is the limit of f(x) as x approaches c from the left or values that are less than c. `color(black)(lim_(xtoc+)f(x))` is the limit of f(x) as x approaches c from the right or values that are greater than c.

For example, in the graph below, the `color(black)(lim_(xtoc)f(x))` does not exist because the `color(black)(lim_(xtoc-)f(x)=2)` and `color(black)(lim_(xtoc+)f(x)=0)`, since 0 does not equal 2, the `color(black)(lim_(xtoc)f(x))` does not exist

Here's some practice.

Evaluating Limits

There are multiple ways to evaluate limits, here are the ways that you will need for the AP test. If one does not work, you can always move on to others. There are also other rules such as L'Hospital's Rule and infinite limits, but those are discussed in another section.

Substitution

Just plug in c into f(x) to find the limit.

Factor and Cancel

If substitution results in 0/0, then one can find a common factor in the numerator and denominator and cancel. Then reevalutate the limit using substitution.

For example, `color(black)(lim_(xto1)(x^2+2x-3)/(x-1))` factors to `color(black)(lim_(xto1)((x+3)(x-1))/(x-1))` which cancels to `color(black)(lim_(xto1)x+3)=4`

Rationalization

If you cannot do a factor and cancel method, but you are still doing 0/0, then you can try rationalization if you have a square root on the top or bottom. It involves taking the inverse of the side that has the square root.

For example, `color(black)(lim_(xto3)(x-3)/(sqrt(4x+4)-4))` results in 0/0, so you can rationalize by multiplying by `color(black)((sqrt(4x+4)+4)/(sqrt(4x+4)+4))` (which is really one so it does not change the expression) to get `color(black)(lim_(xto3)((x-3)(sqrt(4x+4)+4))/(4x+4-16))` which equals `color(black)(lim_(xto3)((x-3)(sqrt(4x+4)+4))/(4x-12))` which equals `color(black)(lim_(xto3)((x-3)(sqrt(4x+4)+4))/(4(x-3)))` which cancels to `color(black)(lim_(xto3)((sqrt(4x+4)+4))/4=2)`

Special Trig Limits

`color(black)(lim_(xto0)(sin(x))/(x)=1)`

`color(black)(lim_(xto0)(1-cos(x))/(x)=0)`

You can memorize these special trig limits, or you can use L'Hospital's so you do not have to memorize them.

One-Sided Limit

When substiution results in 1/0 in a one-sided limit, then the limit equals `color(black)(oo or -oo)`

Here's some practice.

Limit Theorems

Here are the main limit theorems that you need to know for calculus.

`color(black)(lim(kxxf(x))=kxxlimf(x))` where k is a constant

`color(black)(lim(f(x)+g(x))=limf(x) + limg(x))`

`color(black)(lim(f(x)xxg(x)) = limf(x) xx limg(x))`

`color(black)(limf(x)/(g(x))=limf(x)-:limg(x)` where g(x) does not equal 0

`color(black)(limk=k)`

Infinite Limits

`color(black)(lim_(xtooo)f(x))` or `color(black)(lim_(xtooo)f(x))` is the number f(x) approaches as x increases infintely or decreases infintely.

You can use infinite limits to determine horizontal asymptotes. For example:

`color(black)(lim_(xtooo)(x^2-5x+2)/(5x^2+3x+5))` can be determined by taking the highest degree of x and dividing every term by it. So it becomes `color(black)(lim_(xtooo)((x^2)/(x^2)-(5x)/(x^2)+(2)/(x^2))/((5x^2)/(x^2)+(3x)/(x^2)+(5)/(x^2)))`, and whenever you have `color(black)(lim_(xtooo)1/x)` or anything with a higher x degree on the bottom, then the value is automatically zero because it is like doing `color(black)(1/oo)`, which is equal to 0.

`color(black)(lim_(xtooo)((x^2)/(x^2)-(5x)/(x^2)+(2)/(x^2))/((5x^2)/(x^2)+(3x)/(x^2)+(5)/(x^2)))` then becomes `color(black)(lim_(xtooo)(1-0+0)/(5+0+0)=1/5)`

Here's some practice.

Continuity

f(x) is continuous at x=c if:

1. f(c) is defined and c is in the domain. That means that f(c) is a value.

2. `color(black)(lim_(xtoc)f(x))` exists because `color(black)(lim_(xtoc-)f(x)=lim_(xtoc+)f(x))`

3. `color(black)(lim_(xtoc)f(x)=f(c))`

Types of Discontinuities

Your teacher might also say that a function is continuous if you can draw it without having to lift up your pencil. There are three types of discontinuity: jump (jumps from one line to another), removable (has a hole at a point), and infinite (where f(x) approaches infinity or negative infinity so the continuity cannot be determined).

There are also other theorems that rely on continuity

Intermediate Value Theorem: If f(x) is continuous on the closed interval of [a, b] and f(a) < M < f(b) then there must be an x=c between [a, b] where f(c) = M.

Extreme Value Theorem: If f(x) is continuous on the closed interval of [a, b] then there must be a x=c where f(x) is a min or max.

L'Hopital's Rule

If `color(black)(lim_(xtoc)f(x)/g(x)=(+-oo)/(+-oo)` or `color(black)(0/0)` then `color(black)(lim_(xtoc)f(x)/g(x)=lim_(xtoc)(f'(x))/(g'(x)))`

`color(black)((+-oo)/(+-oo), 0/0)` are just two of the indeterminate forms. There are also `color(black)((0)(+-oo), 1^oo, 0^0, oo^0, oo-oo)`. You can use L'Hopital's Rule on those as well, but they are more complex.

Remember, you can only use L'Hopital's Rule if the limit results in one of the indeterminate forms!