Limit+Theorems+and+Properties


 * REMEMBER: SUBSTITUTION SUBSTITUTION SUBSTITUTION.**

If **f** and **g** are functions such that **f(x) = g(x)** for all **x** in some open interval interval containing **a** except possibly for **a**, then



Suppose that **f** and **g** are functions such that the two limits

=Then...=

__LIMIT OF A SUM OF TWO FUNCTIONS__ - "Limit distributes over addition, or the limit of a sum equals the sum of the limits"

f(x) = 2x + 8 g(x) = 4x + 3 a = 2 f(2) = 2(2) + 8 = 12 g(2) = 4(2) + 3 = 11

12 + 11 = 23 (2x + 8) + (4x + 3) = 6x + 11 6(2) + 11 = 23 23 = 23 :)

__LIMIT OF A DIFFERENCE OF TWO FUNCTIONS__ - "Limit distributes over subtraction, or the limit of a difference equals the difference of the limits"

f(x) = 3x + 7 g(x) = x + 2 a = 6 f(6) = 3(6)+ 7 = 25 g(6) = 6 + 2 = 8

25 - 8 = 17 (3x + 7) - (x + 2) = 2x + 5 2(6) + 5 = 17 17 = 17 :)

__LIMIT OF A PRODUCT OF TWO FUNCTIONS__ - "Limit distributes over multiplication, or the limit of a product equals the product of the limits"

f(x) = 8x + 5 g(x) = 2x - 4 a = 5 f(5) = 8(5) + 5 = 45 g(5) = 2(5) - 4 = 6

45 * 6 = 270 (8x + 5) * (2x - 4) = 16x^2 + 10x - 32x - 20 = 16x^2 - 22x - 20 16(5)^2 - 22(5) - 20 = 270 270 = 270 :)

__LIMIT OF A QUOTIENT OF TWO FUNCTIONS__ - "Limit distributes over division, except for division by zero, or the limit of a quotient equals the quotient of the limits"

//Ex. 1// f(x) = 6x + 4 g(x) = 3x + 2 a = 7 f(7) = 6(7) + 4 = 46 g(7) = 3(7) + 2 = 23

46 / 23 = 2 (6x + 4) / (3x + 2) = 2 2 = 2 :)

//Ex. 2// f(x) = 2x + 1 g(x) = x + 3 a = 8 f(8) = 2(8) + 1 = 17 g(8) = 8 + 3 = 11

17 / 11 = 1.5454 (2x + 1) / (x + 3) (2(8) + 1) / (8 + 3) = 17 / 11 = 1.5454 1.5454 = 1.5454 :)

__LIMIT OF A CONSTANT TIMES A FUNCTION__ - "The limit of a constant times a function equals the constant times the limit"

f(x) = 6x + 7 k = 2 a = 3 f(3) = 25 f(k3) = 2(6(3) + 7) = 50 k * f(3) = 2(25) = 50 50 = 50 :)

__LIMIT OF THE IDENTITY FUNCTION (Limit of x)__ - "The limit of x as x approaches a is simply a"

a = 2 Plug in 2 for x. x = 2 2 = 2 :)

__LIMIT OF A CONSTANT FUNCTION__ - "The limit of constant is that constant"

a = 3 k = 5 Since there is no variable to "plug in" a, the limit is just 5.

=**LIMIT OF A PRODUCT OF TWO FUNCTIONS (in case you didn't get enough before)**= Suppose that g(x) = 2x + 1 and h(x) = 5 - x, and f(x) = g(x) * h(x) = (2x + 1)(5 - x) Find the limit of f(x) as x approaches 3. Substitute 3 in for x. f(3) = (2(3) + 1)(5 - 3) f(3) = (6 + 1)(2) = 14 CONGRATULATIONS YOU HAVE FOUND THE LIMIT OF f(x) AS x APPROACHES 3. Here's a medal. Although it looks more like a sun with a one on it.

Let's try another one for some of you mathematically challenged people. Again, f(x) = g(x) * h(x) g(x) = 7x + 2 h(x) = 3 + x Find the limit of f(x) as x approaches 6. f(6) = (7(6) + 2)(3 + 6) f(6) = 53

=LIMIT OF A QUOTIENT OF TWO FUNCTIONS (even more!)= Suppose that g(x) = 2x + 1 and h(x) = 5 - x, and f(x) = g(x) / h(x) = (2x + 1) / (5 - x) f(3) = (2(3) + 1) / (5 - 3) f(3) = 7 / 2 = 3.5

But what if x = 5? That would mean that f(5) = (2(5) + 1) / (5 - 5) f(5) = 11/0 Thus, since there is a zero in the denominator, there is no limit as f(x) approaches 5. the absolute value of the quotient becomes infinitely large and there is a vertical asymptote at x = 5.

=INDETERMINATE FORM= Sometimes even if one follows all of the limit properties and theorems, he/she will end up with 0/0, aka indeterminate form. One way to solve this problem is by factoring. This can be seen in the following problem:

x^3 - x^2 - 2x - 12 / x - 3 as x approaches 3. 3^3 - 3^2 - 2(3) - 12 / 3 - 3 = 0/0 But that answer is just not acceptable. So, to solve this, try to figure out a way to "get rid of" the denominator that becomes 0 when 3 is plugged in. This can be done by factoring the numerator and (hopefully) finding a factor of x - 3

This equation does have a factor of x - 3 So, the new equation becomes (x - 3)(x^2 + 2x + 4) / x - 3 Cancel out both of the (x - 3)'s and one is left with: x^2 + 2x + 4 Now 3 can be plugged in. (3)^2 + 2(3) + 4 = 19 And there is the limit.

I got that one from our math book. But I made up the next one, which is probably why it's extremely easy.

x^2 - x / x - 1 as x approaches 1. Oh, what is this? 0/0? Thankfully your amazing guide (me) taught you how to solve this. The numerator can be factored to (x - 1)(x) / x - 1 Cancel out both of the (x - 1)'s and one is left with just x. Plug in 1, and the limit is 1.

It should also be noted that limx->0 1/x= infinity and is not indeterminate. Simillarly the limit as x approaches infinity of 1/x= 0

Here is a word document that explains the squeeze, mean value and rolle's theorems:

Sources: http://archives.math.utk.edu/visual.calculus/1/limits.18/index.html //Calculus - Concepts and Applications// aka our math book aka most horrible fantastic book in the world.