Suppose we are watching the values of a function f (x) as x approaches x ₀ (without assuming the value of x ₀ itself). (See figure 1) We would like to be able to say that f (x) stays within 1/10 of a unit of L as soon as x stays within some d of x ₀. The ± 1/10 should be treated like an error measurement. So as the error becomes smaller, and smaller, then new d -interval about x ₀ will become smaller. This whole idea leads to the formal definition of the limit.

figure 1

DEFINITION:

Let f (x) be defined on an open interval about x ₀, except possibly at x ₀ itself. We say that f (x) approaches the limit L as x approaches x ₀ and write

if, for every number e > 0, there exists a corresponding number d > 0 such that for all x, 0 < | x - x ₀ | < d Þ | f (x) - L | < e .

EXAMPLE 1:

Find a d > 0 such that for all x 0 < | x - x ₀ | < d Þ | f (x) - L | < e for

SOLUTION: First of all, we need to find the limit of f as x approaches 1.

Now we will work backwards to find d . We will start with | f (x) - L | < e and manipulate it to get | x - 1 | < d .

We want the smaller of the two numbers in absolute value. The reason for this is that it will force the f (x) values to be within the given e value. Which number {-0.4375, .5625} is smaller in absolute value? It is -0.4375, so let d = |-0.4375| = 0.4375.

EXAMPLE 2:

Find a d > 0 such that for all x 0 < | x - x ₀ | < d Þ | f (x) - L | < e for f(x) = 4 - x ² at x ₀ = -1 and e = 0.25.

SOLUTION: First find the limit of f (x) as x approaches -1.

Now work backwards.

EXAMPLE 3:

Find a d > 0 such that for all x 0 < | x - x ₀ | < d Þ | f (x) - L | < e for f(x) = 2x - 2, L = -6, x ₀ = -2, and e = 0.02.

SOLUTION: We are given the value of the limit at x ₀, so let us start working backwards.

So let d = 0.01.

EXAMPLE 4:

Find a d > 0 such that for all x 0 < | x - x ₀ | < d Þ | f (x) - L | < e for f(x) = 1/x, L = 1/4, x ₀ = 4, and e = 0.05.

SOLUTION:

Since -2/3 is the smaller in absolute value, let d = 2/3.

EXAMPLE 5:

Find a d > 0 such that for all x 0 < | x - x ₀ | < d Þ | f (x) - L | < e for

So let d = 0.05.

EXAMPLE 6:

SOLUTION: To prove this limit we must find a d > 0 such that for all x, 0 < | x - 3 | < d Þ |(3x - 7) - 2 | < e where e > 0. To do this, we will work backwards as before.

EXAMPLE 7:

SOLUTION: To prove this limit we must find a d > 0 such that for all x, 0 < | x - 9 | < d Þ

Why did I pick Abs((2 - e ) ² - 4) for d ? It is the smaller of the two values in absolute value.

EXAMPLE 8:

SOLUTION: To prove this limit we must find a d > 0 such that for all x, 0 < | x - 1 | < d Þ