The graph of f(x). Here, we want to find
We'll call the limit "L" (I happen to know the limit is 1.6)
By the definition of "limit", if you give me any ε, I should be able to come up with a δ such that
Whenever x is within δ of 4, f(x) will be within ε of L
We start with ε , and we show the allowable values of f(x)
We will try to keep the blue dots representing f(x) between the red lines by choosing a δ that restricts the values of x.
Here, we choose a δ that is small enough so that whenever x is within δ of 4 -- that is, whenever x is not equal to 4 and between the green lines, f(x) will be between the red lines.
In other words, there won't be any function values (the blue line) that are between the green lines but not between the red lines.
As you can see from this diagram, the test is satisfied for this choice of ε and δ. Also, you should be able to see that if the slope is very high, then δ must be very small, compared to ε, to make sure the graph doesn't pierce roof or floor of the rectangle formed by the red and green lines.
1. The form of the statement you will need to prove is
2. Your main job in proving a limit is to find an expression for δ in terms of ε that satisfies the definition of the limit (for any ε there exists a δ such that
if 0 < |x - c| < δ then |f(x) - L| < ε
3. There's no penalty for picking a δ that is too small; if anything, it makes the statement easier to prove. So start by setting an upper limit on the size of δ -- say, δ < 1.
4. Now, find the maximum (in absolute value) slope, m, of all the lines that goes through (c,L) and a point on the graph between c-1 and c+1. Again, there's no penalty for picking a slope, m, that's too big, so overestimate the slope if it makes your job easier.
5. Now write the expression for δ as follows:
δ = min(1,ε/m)
6. Finally, prove the "if-then" statement that was given in step 2. Start with the "if" part, and substitute min(1,ε/m) in place of δ as a first step in proving the "then" part.
This example uses the tips, above, to prove a particular limit. The numbers of the steps correspond to the tip numbers, to make it easier for you to follow the example.
1. Prove limx−>2 x² = 4
2. We will prove the following statement:
if 0 < |x - 2| < δ then |x² - 4| < ε
3. The δ we will pick will be in no circumstance larger than 1. So we need be concerned only with the interval (1,3).
4. Let m be the absolute value of the slope of a line that goes through (2,4) and a point (x,x²) where 1 < x < 3.
m = |(x²-4)/(x-2)|
Now, by dividing numerator and denominator by x-2, we get
m = |x+2|
The maximum slope of any such line on the interval (1,3), then, is m=|3+2|, or 5.
5. Now, we will write the equation for δ as
δ = min(1,ε/5)
6. Now, I will start again with the "if-then" statement from step 2, assuming the "if" part, and work to prove the "then" part:
0 < |x - 2| < δ
Substituting min(1,ε/5) in place of δ,
|x - 2| < min(1,ε/5)
Since |x-2| is less than the min of two things, it is less than each one:
|x - 2| < 1 and |x - 2| < ε/5
Since |x-2| < 1, it follows that |x+2| < 5. Then I multiply both sides of |x-2|<ε/5 by |x+2|
|x-2||x+2| < (ε/5)|x+2| < (ε/5)(5) = ε
The leftmost side is less than the rightmost side, so we have the final result,
|x² - 4| < ε
Introduction to limits
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