## 1. Infinite Limits

### 1.1. Replace with "x+0.001" or "x-0.001", then solve. While solving if answer is positive then final answer = positive infinite, if answer negative then final answer = negative infinite.

## 2. Limits to Infinite

### 2.1. a=b) Lim x->inf M/N

2.1.1. (3x^2 + 5x -2)/(2x^2 + 7x -7) look for the bisggest exponent everything else with a lower exponent will become a zero (It'll cancel) and use its coefficient as "a" and "b" 3/2

### 2.2. a>b) Lim x-> UNDFINED

2.2.1. (3x^3 + 5x -2)/(2x^2 + 7x -7) look for the bisggest exponent everything else with a lower exponent will become a zero (It'll cancel) and use its coefficient as "a" and "b"(3/0) = UNDEFINED

### 2.3. a<b) Lim x-> 0

2.3.1. (3x^2 + 5x -2)/(2x^3 + 7x -7) look for the bisggest exponent everything else with a lower exponent will become a zero (It'll cancel) and use its coefficient as "a" and "b" (0/2) = 0

### 2.4. a = The biggest coefficient in the numerator

### 2.5. b = The biggest coefficient in the denominator

### 2.6. Always look for the biggest exponent in the variable then use a and b

## 3. For a limit to exist it ahould:

### 3.1. Lim must exist x=a

### 3.2. Lim must exist x>a+

### 3.3. Lim must exist x>a-

### 3.4. Lim x>a+ = Lim x>a- = Lim x>a

## 4. Grapically

### 4.1. If "x" tends to "#", then "y" tends to "#"

4.1.1. If you have a graph and given value of "x" just search for the "y" value using the rule in the bottom.

## 5. Numericallly

### 5.1. X+0.001 and X-0.001, Simplify then replace, solve, then get the average.

5.1.1. (x^2 +2x -35)/(x-5) when Lim x->5 use 4.999 and 5.001 as values for x. Simplify the equation and then replace (x+7) = (4.999+7) = 11.999 (5.001+7) = 12.001 Get the average 12

## 6. Algebraically

### 6.1. First simplify. then replace and solve.

6.1.1. (x^2 +2x -35)/(x-5) when Lim x->5 simplify equation (x+7) then replace for the valur of "x" (5+7) = 12