Limits from a graph

A graph with a hole

Picture the line f(x) = x + 2, but with one point punched out at x = 2. The point (2, 4) is missing: that's a hole (an open circle on the graph). The function has no value of 4 at x = 2, yet everything around that spot clearly lines up with 4.

Sneaking up from both sides

To see where the graph is heading, try x-values closer and closer to 2:

| x (from the left) | f(x) | x (from the right) | f(x) | |---|---|---|---| | 1.9 | 3.9 | 2.1 | 4.1 | | 1.99 | 3.99 | 2.01 | 4.01 |

From the left, y creeps up toward 4. From the right, y comes down toward 4. These are the one-sided limits: the left-hand limit lim x→2⁻ f(x) = 4 and the right-hand limit lim x→2⁺ f(x) = 4.

When both sides agree

The (two-sided) limit exists only when both one-sided limits exist and are equal. Here they're both 4, so

lim x→2 f(x) = 4

Read it as "the limit of f of x, as x approaches 2, is 4."

Heading to vs landing on

Now suppose someone defines f(2) = 1, a single dot down at (2, 1), below the hole. Does the limit change? No. The limit is still 4, because it only looks at x-values near 2, never at x = 2 itself. The function's value f(2) = 1 and the limit 4 are two different questions: where y lands and where y is heading.

Worked example

Find lim x→1 h(x) for h(x) = 2x with a hole at x = 1.

• From the left: x = 0.9 gives 1.8; x = 0.99 gives 1.98.
• From the right: x = 1.1 gives 2.2; x = 1.01 gives 2.02.
• Both sides head to 2, so lim x→1 h(x) = 2, even though h(1) isn't defined.

Try it yourself

k(x) = x − 1 has a hole at x = 4, and someone sets k(4) = 10. Make a small table with x = 3.9, 3.99, 4.1, 4.01. What is lim x→4 k(x), and what is k(4)? (Answer: the limit is 3, because the table gives 2.9, 2.99, 3.1, 3.01; but k(4) = 10. Different questions, different answers.)

Good to know

• This is the intuitive (graph and table) idea of a limit used at the start of AP Calculus AB/BC; the formal epsilon-delta definition comes later.
• The two-sided limit exists only when the left-hand and right-hand limits exist and are equal.
• In the example, f(x) = x + 2 with the point at x = 2 removed and redefined as f(2) = 1. That's a removable discontinuity, the topic of the next lesson, Continuity.
• A hole means the point (2, 4) is not on the graph. The limit never depends on the value at the point itself.

Answer to the comment question

g(x) = x + 5 has a hole at x = 3. What's the limit as x → 3? The limit is 8. Near x = 3 the graph follows x + 5, which heads to 3 + 5 = 8 from both sides (2.99 + 5 = 7.99, 3.01 + 5 = 8.01). The hole doesn't change the limit.