Calculus Calculating Limits
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There's Got to be a Limit
Calculating Limits

 By now, you have seen equations like these:
Y = X2 - 3X + 5
F(X) = X2 - 3X + 5
The difference between these two is just picky details.
Just think of them as the same.
In these pages, we will often talk about F(X) or G(X) or something like it.
This is just a way of saying: "some equation where X is the variable."
Like the X 2 - 3X + 5 stuff we just had.
 If we talk about F(t) or G(t), that just means we are talking about some stuff
 where the variable name is t.
 Like maybe:

 F(t) = 16t3 - 4t2 + 2 

When we solve a function for some X (or t) value,  it's usually no big deal.
We just put the number into the equation in place of the variable
 and work out the arithmetic.
If we wanted to solve the function above for t = 2, we just put in 2 for t. 

 F(t) = 16t3 - 4t2 + 2

 F(t) = 16(2)3 - 4(2)2 + 2

 F(t) = 16 x 8 - 4 x 4 + 2

 F(t) = 128 - 16 + 2

 F(t) = 114

 No biggie, eh?
 What if we have this:

F(x) = 


First, we can use algebra to simplify it:
F(x) =  X2 x X 1

X 1

F(x) = X2

 Maybe you remember this function from algebra as a parabola. 
The graph of the function looks like this:


 BUT, we have a problem.
 Our original function had a denominator, X.
 When the denominator equals ZERO,
 the function will not work out to some nice value: 
 Putting 0 in for X in our original function, we get:






 And THAT puppy is undefined.
 We can't deal with a zero in the denominator...
 If we graphed the original function, we get:

X - 1 - 0.7 - 0.3 - 0.2 - 0.1 0 0.1 0.2 0.3 0.7 1
F(X) 1.00 0.49 0.09 0.04 0.01 undef 0.01 0.04 0.09 0.49 1
 Up to now, if a denominator on one side of a function was zero,
 we just said that the function was undefined.
 Are we saying that there is a way to get around that zero in the denominator
 and find an answer?
 Differential Calculus could also be called: 
 "How to get an answer when you have a zero in the denominator"
 Here's how it works.
 If you try to solve an equation for a value of X that makes the denominator zero,
 you get nowhere fast.
 You CAN solve the equation for values of X
 a tiny bit bigger or smaller than the problem value.
 We want to see what happens to the value of the function
  as we get really, really, close to the value of X 
 that gives us the denominator zero.
 Sometimes, the value of the function gets closer and closer to some number.
 That's what happened in our example.
 As X got closer and closer to the problem value,
 the function got closer and closer to a value.
 It just so happened in this case,
 that both the problem X value and the function value
 we got closer and closer to were the same number. Zero.
 It usually doesn't happen that both numbers are the same.
 Here's the big leap to calculus...
 We say:
 If the closer we get to some undefined X value,
 the closer the function value gets to some number.
 And the function value we get closer to is the same
 if the close X value is bigger or smaller than the undefined X value.
 That's Calculus!
 It's basically saying:
 Yeah I KNOW that the function is undefined at that X value,
 but if it wasn't what would the value be.
 It's an amazing thing. 
 If you said something like that in Algebra,
 the teacher would have probably smacked you around. 
 With words, anyway.
 But it's the guts of Calculus.
 Math types have invented all kinds of fancy symbols and terms
 to try to hide behind.
 But that's the deal.
 Differential Calculus is division where the denominator is zero
 and we use the little "What would the answer have been" scam
 to get the answers.
 The same scam makes Integral Calculus work too, but we get to that later.
 The scam has an official math name.
 It's called Finding the Limit.
 They say:
 We are finding the limit as X approaches
 (the value that makes the function undefined).
 Find out what the function value is when X = 2 if it wasn't undefined for ...
 Oh wait, we're in calculus now
 OK, Find the limit of F(X) as x approaches 2 for the following function:
F(X) X2(X-2)

x 1.80 1.90 1.95 1.99 2 2.01 2.05 2.10 2.20
F(x) 3.24 3.61 3.80 3.96 undef 4.04 4.20 4.41 4.84
 So when X equals 2 we have big problems.
 The denominator equals zero and the function is undefined.
 But it sure looks like the closer X gets to 2, the closer the function gets to 4
 Two things are really important here.
 One is that the function value gets closer and closer to 4
 if X is bigger than 2 or if X is smaller than 2.
 The other is that the value the function is getting closer to
 is an actual number, not infinity.
 If those two things are true,
 same function value with X values bigger and smaller than the problem value.
 And the function value we are closing in on is a number, not infinity.
 We get to use the scam and say that is the Limit of the function
 as X approaches the problem value.
 And treat that function value as if it were the answer we would actually calculate.
 If one or the other of the two things are not true,
 math types say the limit does not exits.
 That means we're stopped again.
 At least for now.
 Of course math types invented notation for this.
 If you wanted to find the limit of something as X approached 2, you would write:
 in front of the thing.
 Lim is the abbreviation for Limit
 WOW We saved two whole letters!
 So for the example we did,
 we would take the limit of both sides as X approached 2.
 We would write:
Lim F(X) Lim X2(X-2)

X 2 X 2 (X-2)
 Now just simplify:

 F(X) = 

Lim X2
X 2 X 2
 And if the zero problem is gone from the denominator,
 substitute in the limit value for X and solve.

F(2) = 22

F(2) = 4

 Here's the graph of the whole thing:

 If we can't factor away the terms that will make the denominator equal to zero,
 we're stopped.
 The limit does not exist.
 If we can factor those terms away,
 the next step is to put the "Lim" notation up next to the function.
 Lim is very powerful.
 You can use it with any type of math function there is.
 Yes it slices, it dices, it makes mounds and mounds of julienne fries
 and it's not available in any stores.
 Lim is also kind of blind.
 It doesn't notice anything in a function except the variable
 named right below the Lim.
 If we had the expression:

3X + 2Y + 5Z

 and we wanted to find the limit as Z approached 3:
Lim 3X + 2Y + 5Z
Z 3
 The Lim will ignore the X and the Y and go right for the Z.

3X + 2Y + 5



Z 3
 It substitutes the number below it for Z

 3X + 2Y + 5 x 3

 and simplifies the expression as much as possible.

 3X + 2Y + 15


   copyright 2005 Bruce Kirkpatrick

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