Download Finding δ Given a Specific ϵ - Examples

Finding δ Given a Specific - Examples
1. Find δ given the following information:
f (x)
x0
L
√
=
x−1
= 5
= 1
= 2.
We are trying to formally show that
lim
x→5
√
x − 1 = 2.
The value for we are given is telling us how close to our limit, L, we want to be. Our goal
is to find a value for δ so that if we stay within the “δ-neighborhood” of x0 , the value of the
function will be in the “-neighborhood” of L.
In case you have never heard these terms before, here is what they mean:
A “δ-neighborhood” around a point x0 is a small interval around x0 . To find it, you simply add
δ to x0 (to get the right endpoint), and subtract δ from x0 (to get the left endpoint). Remember
that δ is just a very small number, like 0.1. The result is an interval where the left endpoint is
x0 − δ, the right endpoint is x0 + δ, and x0 is in the center. Try graphing that on a number line
to see if it makes sense. The same thing is true for an “-neighborhood”.
Back to our problem. We need to go through the steps we talked about in class to find δ.
Step 1: Solve |f (x) − L| < to look like a < x < b. (a and b will just be numbers)
√
| x − 1 − 2|
√
−1 < x − 1 − 2
√
1< x−1
1<x−1
2<x
<
<
<
<
<
1
1
3
9
10.
Now we can see that for our problem, a = 2 and b = 10. This tells us what values x has to be
in between in order for our function, f (x) to stay within 1 unit (that’s ) of 2 (that’s L).
Step 2: Solve |x − x0 | < δ to look like −δ + x0 < x < δ + x0 . Then solve a = −δ + x0
and b = δ + x0 for δ, and choose the smallest!
|x − 5| < δ
−δ < x − 5 < δ
−δ + 5 < x < δ + 5.
Next, we need to set up the equations, putting both parts together:
−δ + 5 = 2
−δ = −3
δ = 3,
and
δ + 5 = 10
δ = 5.
Choose the smallest of the two. So, δ = 3! We’re done!
2. Find δ given the following information:
f (x)
x0
L
=
=
=
=
x2
√
3
0.1
3.
We will use the same steps as in #1 to again find δ.
Step 1:
|x2 − 3|
−0.1 < x2 − 3
2.9 < x2
√
2.9 < x
So a =
√
2.9 and b =
√
<
<
<
<
0.1
0.1
3.1
√
3.1.
3.1.
Step 2:
√
|x − 3| < δ
√
−δ < x − 3 < δ
√
√
−δ + 3 < x < δ + 3.
Now set up the equations, again putting both parts we’ve done together:
√
√
−δ + 3 =
2.9
√
√
−δ =
3 − 2.9
δ = 0.032,
and
δ+
Therefore, δ = 0.028! Done!
√
√
3 =
3.1
δ = 0.028.