Continuity of Functions Exercises

Answer

Since c = 3,

f(c) = f(3) = 5(3) – 2 = 13.

Now we turn to the inequality f(c) – ε < f(x) < f(c) + ε, which for this function is, and value of ε is

13 – 1 < 5x – 2 < 13 + 1

and solve for x:

12 + 2 < 5x < 14 + 2, so

2.8 < x < 3.2

Now we can finish it up by subtracting c from all parts of the inequality to find δ.

2.8 – 3 < x – 3 < 3.2 – 3

-0.2 < x – 3 < 0.2

so δ = 0.2.

Answer

We start with the inequality f(c) – ε < f(x) < f(c) + ε and fill in what we're given for c, f, and ε:

e⁰ – 1 < e^x < e⁰ + 1,

or

0 < e^x < 2.

Now use the natural log to solve for x:

ln(0) < ln(e^x) < ln(2)

ln(0) < x < ln(2)

Here we have a bit of a problem, though. We can't find ln(0); it doesn't exist. What we can say though, is that since ln(x) has approaches -∞ as x approaches 0,

-∞ < x < ln(2).

Next is where we would usually subtract c from the entire inequality. Since c = 0, that's not going to change the inequality, though, and we can just say δ is ln(2).

Answer

We're getting the hang of this now. We always start with the inequality

f(c) – ε < f(x) < f(c) + ε

and fill in what we were given for c, f, and ε:

6 – 0.2 < f(x) < 6 + 0.2, so

5.8 < x² + 2 < 6.2.

Solve the inequality for x. Subtract 2 from all parts of the inequality to find

3.8 < x² < 4.2, then take square roots and round to find 1.949 < x < 2.049.

Notice this doesn't make sense. x is supposed to be close to -2, not close to 2. The problem is that we need to take negative square roots, which gives us

-1.949 > x > -2.049

(also notice that the direction of the inequalities need to switch, since -1.949 is bigger than -2.049). 

Finally, subtract c from all parts of the inequality and we'll have δ. In this case c = -2, so

-1.949 – (-2) > x – (-2) > -2.049 – (-2)

0.051 > x - (-2) > -.049.

Therefore x needs to be within 0.049 of -2, so

δ = 0.049.