The Formal Version Examples

Let f(x) = 2x, c = 4, and ε = 0.5. Find an appropriate δ > 0, so that if x is within δ of 4, f(x) will be within ε of f(c).

This is the same problem as an earlier example, but we're going to solve it a different way.

First, since c = 4, f(c) = 8. 

Saying we want f(x) to be within ε of 8 means we want to have 7.5 < f(x) < 8.5, or 7.5 < 2x < 8.5. We can solve this inequality in one step: divide through by 2 to find 3.75 < x < 4.25. 

Now, subtract 4 from all parts of the inequality: 3.75 - 4 < x – 4 < 4.25 – 4 and -0.25 < x – 4 < 0.25. This means x must be within 0.25 of 4, so δ = 0.25.

Let f(x) = x², c = 2, and ε = 0.1. What is an appropriate value of δ > 0, that will guarantee that if |x – c| < δ, |f(x) – f(c)| < ε?

• The first thing we want to do is set up an inequality that looks like
  f(c) - ε < f(x) < f(c) + ε.
  In this case, we have c = 2 so f(c) = 2² = 4.
  Since ε = 0.1, we find 4 – 0.1 <  f(x) < 4 + 0.1, or 3.9 < x² < 4.1.

• Now we need to solve the inequality.
  Taking square roots of all parts of the inequality, we find  Evaluating the square roots and rounding a lot, we need to have 1.97 < x < 2.02.

• To find δ, subtract c = 2 from all parts of the inequality: 0.03 < x – 2 < 0.02. This says x can move to 0.03 away from 2 to the left, but only 0.02 away from 2 to the right. Since we want to restrict the values of x so c = 2 is in the middle, we choose the smaller bound, and go with δ = 0.02. Now we can check this answer. As long as x is within δ of c, that is, 2 – 0.02 < x < 2 + 0.02, we find (1.98)² < x² < (2.02)². Then 3.9 < 3.9204 < x² < 4.0804 < 4.1, which means 3.9 < f(x) < 4.1. That's just what we wanted.

Let f(x) = 4x + 1. 

Find a value of δ such that if |x – 5| < δ, then |f(x) – f(5)| < 0.1.

Since we're looking at how far x moves from 5, and how far f(x) moves from f(5), we have c = 5. Therefore

f(c) = f(5) = 4(5) + 1 = 21.

Since we want |f(x) – f(5)| < 0.1, we have ε = 0.1.

Now we follow our recipe.

• We write down the inequality f(c) – ε < f(x) <  f(c) + ε. Now fill in the known variables: 21 – 0.1 < f(x) < 21 + 0.1 so 20.9 < 4x + 1 < 21.1

• Solve the inequality for x.
  Subtract 1 from each part of the inequality to find 19.9 < 4x < 20.1, then divide each part by 4 to find 4.975 < x < 5.025.

• Subtract c from all parts of the inequality to find δ.
  Since c = 5, we subtract 5 from each part of the inequality to find -0.025 < x – 5 < 0.025 therefore |x – 5| < 0.025. Then δ = 0.025 is the number we want.

We can check our answer by working backwards:

If |x – 5| < 0.025, then

-0.025 < x – 5 < 0.025 and 4.975 < x < 5.025.

Now multiply through by 4 and add 1 to find 19.9 < 4x < 20.1, then 20.9 < 4x + 1 < 21.1

Since f(x) = 4x + 1, we can rewrite this as 20.9 < f(x) < 21.1.

Subtract f(5) = 21 from each part to find 20.9 – f(5) < f(x) – f(5) < 21.1 – f(5)

-0.1 < f(x) – f(5) < 0.1

Therefore |f(x) - f(5)| < 0.1, which was what we wanted to begin with.

Let f(x) = sin(x) 

Find a value of δ such that if , then .

We start with ,   , and ε = 0.005.

• Now write down the inequality
  f(c) – ε < f(x) <  f(c) + ε
  and fill things in: 0.5 – 0.005 < sin(x) < 0.5 + 0.005, so 0.495 < sin(x) < 0.505.

• Solve the inequality for x.
  To do this, take the inverse sine of each piece of the inequality: arcsin(0.495) < x < arcsin(0.505).

Then evaluate and round each arcsin: 0.517835 < x < 0.529382.

• Subtract c from all parts of the inequality and find δ.
  Subtracting  and rounding some more, we find .

The value 0.0057 is less than both of these bounds, therefore we can take δ = 0.0057.

Let . Find a value of δ such that if |x – 2| < δ, then |f(x) – 0.5| < 0.005.

We have c = 2 and ε = 0.005. Now we'll cook up δ.

• Start by writing down the inequality, f(c) – ε < f(x) < f(c) + ε, then fill things in:                                                                              

• Now solve the inequality for x.
  Multiplying through by x we find 0.495x < 1 < 0.505x.

This is equivalent to having the two inequalities

0.495x < 1
1 < 0.505x

Solving these inequalities, we find

 rounding a little,

1.98 < x < 2.02.

• Subtract c from all parts of the inequality to find δ.
  Subtracting 2, we find -0.02 < x – 2 < 0.02, so take δ to be 0.02.

Let . What is a value of δ that will guarantee that if |x – 7| < δ, then ?

We're starting with c = 7, , and ε = 0.1. Now we'll find a recipe for δ.

• As always, the next step is to write out the inequality f(c) – ε < f(x) < f(c) + ε and fill things in:                                                     
•                                                               

• Next we solve the inequality for x.
  Square each part of the inequality to get 
  And now subtract 3 to solve for x:
•       

• To finish up, we'll subtract c from all parts of the inequality to find δ.
  
• Our value for δ is  which is about 3.64.