Limits: what happens near a point

TL;DR:

• A limit asks what value f(x) approaches as x gets close to a point, not what happens at the point.

• When substitution gives 0/0, simplify by factoring and cancelling, then substitute.

• A limit exists only if the left-hand and right-hand values agree.

Take f(x) = (x² − 1)/(x − 1). Put in x = 1 and you get 0/0, which has no value. Yet the function behaves nicely just next to 1. Limits are the tool for describing that: what the function is heading toward, even where it is not defined.

Close to a point, not at it

The limit of f(x) as x → a is the number f(x) gets arbitrarily close to when x is close to a, from either side, without caring what happens at x = a itself. The function might not exist at a, or might have a different value there, and the limit does not change.

Worked example: (x² − 1)/(x − 1) as x approaches 1

Start with a table of values on both sides of 1.

Try one input by hand. With x = 0.9 the numerator is 0.81 − 1 = −0.19 and the denominator is −0.1, giving −0.19 / −0.1 = 1.9. Come in from either side and the outputs settle toward 2.

Now explain why with algebra. Factor the top as a difference of squares: x² − 1 = (x − 1)(x + 1). For any x ≠ 1, the factor x − 1 cancels with the bottom, leaving x + 1. So the function equals x + 1 everywhere except at x = 1, where it has a hole. The limit as x approaches 1 is 1 + 1 = 2.

Why cancelling is allowed

The limit only looks at x values near 1 but not equal to 1. For all of those, x − 1 is not zero, so dividing top and bottom by it is legal. The cancelled function x + 1 and the original agree at every point the limit examines. So they have the same limit, even though they differ at the single point x = 1.

Try two more yourself

First, lim x→3 (x² − 9)/(x − 3). Substituting gives 0/0. Factor x² − 9 = (x − 3)(x + 3), cancel, and substitute: 3 + 3 = 6. Table check: at x = 2.9, (8.41 − 9)/(−0.1) = 5.9; at x = 3.1, (9.61 − 9)/0.1 = 6.1. Both sides head for 6.

Second, lim x→2 (x² − 5x + 6)/(x − 2). Factor the top as (x − 2)(x − 3), cancel x − 2, and substitute: 2 − 3 = −1. Check at x = 2.01: the top is 4.0401 − 10.05 + 6 = −0.0099, and dividing by 0.01 gives −0.99. It heads for −1.

The first of these is the derivative of x² at x = 3, in disguise: ((3 + h)² − 9)/h = 6 + h, which tends to 6 as h shrinks. This is how limits become slopes.

One-sided limits

Sometimes the left and right values disagree. Consider g(x) = |x| / x. For positive x, |x| = x, so g = 1. For negative x, |x| = −x, so g = −1. As x approaches 0 from the right, the limit is 1. From the left, it is −1. Because the two sides disagree, the limit at 0 does not exist.

Notation helps: lim x→0⁺ means from the right and lim x→0⁻ from the left. A two-sided limit exists only when both one-sided limits exist and are equal. In our main example, both sides give 2, so the limit is 2.

Other ways to evaluate a limit

• Direct substitution. If the function is continuous at the point, just plug in. For x² + 3 as x → 2, the limit is 4 + 3 = 7.

• Factor and cancel. For 0/0 forms that come from polynomials.

• Tables or a graph. A good way to see the trend, but not a proof.

• Rationalize. For roots, multiply top and bottom by a conjugate.

How to check your answer

• Table. Evaluate at several points on both sides and see if they agree.

• Graph. Sketch x + 1 with a hole at (1, 2).

• Substitute into the simplified form. 1 + 1 = 2.

• Compare sides. Do the left and right limits match?

Common mistakes

• Thinking that 0/0 means the limit does not exist. It means more work is needed.

• Cancelling terms rather than factors. In (x² − 1)/(x − 1), you cannot cancel the x's.

• Confusing the limit with the value of the function at the point.

• Only checking from one side.

• Substituting before simplifying and giving up.

Where you will use it next

Limits are the foundation for the derivative, where the slope is the limit of (f(x + h) − f(x))/h as h → 0, and for the fundamental theorem of calculus. Limits are what make the slope of a curve at a point meaningful, since a single point has no width to measure a rate over.

Frequently asked questions

Can a limit exist where the function is undefined?

Yes. Our example is undefined at 1, yet its limit there is 2.

Does a limit have to equal the function value?

No. When they are equal, the function is called continuous at that point. In our example, they do not match, since the function has no value at 1.