Limits: what they are, how to find them from a table, a graph and the algebra, the limit laws, and what to do when you get 0/0

The limit of f(x) as x approaches a is the value f(x) gets close to as x gets close to a, from both sides, without x ever being a. It is not f(a). That is why a limit can exist where the function is undefined, and why substituting and getting 0/0 is not an answer but an instruction to do more algebra. This page finds one limit three ways, states the limit laws, and shows the two algebra moves (factoring, rationalising) that turn 0/0 into a number.

What a limit is in calculus, and what it is not

A limit is a statement about a function near a point, not at it. Write lim as x → a of f(x) = L and you are saying: as x gets closer and closer to a, from the left and from the right, f(x) gets closer and closer to L. The value f(a) is not part of the claim. The function may equal L there, may equal something else, or may not be defined at a at all, and the limit is unchanged.

"From both sides" matters. A one-sided limit looks only from the left (x → a⁻) or only from the right (x → a⁺). The two-sided limit exists only when the two one-sided limits exist and agree. A function with a jump at a has a left limit and a right limit that differ, so it has no two-sided limit there.

Limits are the first idea of every calculus course because the derivative is a limit, and so is the integral. Everything after this page rests on it.

Numerically: the table of values

One example runs through this whole page: g(x) = (√x − 2)/(x − 4) as x → 4. Substituting x = 4 gives (2 − 2)/(4 − 4) = 0/0, so the function is undefined there. Here is what it does nearby, each value to five decimal places.

x	g(x)
3.9	0.25158
3.99	0.25016
3.999	0.25002
4.001	0.24998
4.01	0.24984
4.1	0.24846

From the left the values come down toward 0.25 from above; from the right they come up toward 0.25 from below. Both sides close in on the same number, which is strong evidence that the limit is 1/4.

A table can only be evidence. It cannot prove the limit is exactly 1/4 rather than 0.2500001, and it cannot see what happens closer to 4 than its last row. That is what the algebra is for. But a table is the right first move when you do not yet know what the function does, and a table whose two sides disagree is a warning, before you spend a line on algebra, that there may be no limit.

Graphically: the hole

The graph of g is the curve y = 1/(√x + 2), as the algebra below will show, with one point missing: there is a hole at (4, 1/4). The curve comes in from the left at a height just above 1/4, skips the single point where x = 4, and continues on the right at a height just below 1/4. The limit is the height of the hole.

Reading a limit off a graph is the same question: as you trace toward x = a from the left and from the right, what height are you approaching? If both traces head for the same height, that is the limit, whether or not a dot sits there. If the graph jumps, the two traces head for different heights, and there is no two-sided limit.

Algebraically: how to find limits in three steps

1. Substitute. Put a in for x. If you get a number, you are done: that number is the limit, for polynomials, roots and rational functions whose bottom is not zero there.
2. If you get 0/0, factor or rationalise. 0/0 means the top and the bottom share a factor that is zero at a. Find it. For polynomials, factor; for roots, multiply top and bottom by the conjugate.
3. Cancel the shared factor and substitute again. The cancelling is allowed because x is near a, never equal to a, so the factor is never zero where the limit looks.

If substitution gives a nonzero number over 0, the function blows up near a: the limit is infinite, or the two sides go opposite ways and the limit does not exist. Check the sign of the bottom on each side to say which.

Worked example: rationalising, line by line

Find lim as x → 4 of (√x − 2)/(x − 4).

• Line 1, substitute. (√4 − 2)/(4 − 4) = (2 − 2)/0 = 0/0. Not an answer; go to step 2.
• Line 2, multiply top and bottom by the conjugate of the top, √x + 2. ((√x − 2)(√x + 2))/((x − 4)(√x + 2)). Leave the bottom factored.
• Line 3, expand the top. (√x − 2)(√x + 2) = x − 4, a difference of squares.
• Line 4, cancel. (x − 4)/((x − 4)(√x + 2)) = 1/(√x + 2), allowed because x ≠ 4 near the limit.
• Line 5, substitute. 1/(√4 + 2) = 1/(2 + 2) = 1/4.

The algebra says exactly 1/4, and the table above was closing in on 0.25. Three methods, one answer. Say the limit is 1/4, not 0.25: the fraction is the exact value and the decimal is what the table approached.

The common wrong lines

"0/0 = 0", or "0/0 is undefined, so the limit does not exist." Both read 0/0 as a result. It is not; it is a signal that the question is still open. The table shows the limit is 1/4, which is neither 0 nor nonexistent. 0/0 is called an indeterminate form for this reason: different functions that give 0/0 at a point have different limits there, so the form alone tells you nothing.

Multiplying by the conjugate on the bottom only, or expanding the bottom. The conjugate must go on top and bottom, or the function has changed. And if you expand (x − 4)(√x + 2) into x√x + 2x − 4√x − 8, the factor x − 4 disappears from sight and the cancellation that was the whole point cannot happen. Leave the bottom factored until the shared factor is gone.

Factoring, the other move

When the function is a ratio of polynomials, the shared factor is found by factoring. Find lim as x → 2 of (x² − 5x + 6)/(x − 2). Substituting gives (4 − 10 + 6)/0 = 0/0. Factor the top: x² − 5x + 6 = (x − 3)(x − 2). Cancel the x − 2: the function equals x − 3 near 2. Substitute: 2 − 3 = −1. The limit is −1. Same three steps, a different second move.

A sibling to try cold

Cover the worked example and find lim as x → 9 of (√x − 3)/(x − 9). Write the substitution line first, even though you know it will be 0/0; the line is the hypothesis check that tells you step 2 is needed.

Check

1/6. After rationalising with √x + 3, the top is x − 9, the x − 9 cancels, and 1/(√x + 3) at x = 9 is 1/(3 + 3) = 1/6.

If you got 0, you read 0/0 as an answer. If your conjugate was √x − 3, you multiplied by the same thing and the root did not clear.

The limit laws

The laws say that limits behave well with arithmetic, each in one line.

• The limit of a sum or difference is the sum or difference of the limits.
• The limit of a product is the product of the limits, and a constant multiple comes out in front.
• The limit of a quotient is the quotient of the limits, provided the bottom's limit is not 0.
• The limit of a power or a root is the power or root of the limit (for an even root, the limit inside must not be negative).

Together, the laws are why "substitute" works. A polynomial is sums and products of x and constants, and the limit of x as x → a is a, so the limit of a polynomial at a is its value there. A rational function is a quotient of two polynomials, so the same holds wherever the bottom is not zero.

The quotient law's condition is exactly where step 2 lives. For lim as x → 3 of (x² − 9)/(x − 3), the bottom's limit is 0, so the quotient law says nothing. Factor: (x − 3)(x + 3)/(x − 3) = x + 3 near 3, and the limit is 6. The laws give the easy cases; factoring and rationalising give the ones the laws refuse.

Continuity, in three conditions

A function f is continuous at a when three things hold: f(a) exists; the limit as x → a exists; and the two are equal. Each condition can fail on its own.

• The hole at x = 4 in this page's example fails the first condition: the limit is 1/4, but g(4) is not defined.
• A jump fails the second: the two one-sided limits disagree, so there is no limit to compare with f(a).
• A function that is defined at a, with a limit at a, but with f(a) set to some other value fails the third.

For a Calculus I reader there is one more sentence: the Intermediate Value Theorem says a function continuous on a closed interval [a, b] takes every value between f(a) and f(b) somewhere in the interval. It is the reason a continuous function that is negative at one end and positive at the other has a zero in between, and it is idea 3 of the first-year course.

Where this sits in the course, and where it goes next

In MCV4U limits are three ideas in a row: idea 2, Limits of functions (numerically, graphically, algebraically), which is this page's three sections; idea 3, Limit laws and evaluating limits by factoring and rationalising; and idea 4, Continuity. Before them is idea 1, Average and instantaneous rates of change, and after them is idea 5, The derivative from first principles. The MCV4U course map shows all 24 ideas in teaching order.

In a first-year Calculus I course this is idea 1 of 26, Limits of functions and limit laws, and idea 3, Continuity and the Intermediate Value Theorem. Between them sits idea 2, the formal ε-δ definition of the limit, which this page does not cover. First-year calculus, what changes is written for the Grade 12 student about to meet it, and is calculus hard answers the question most students ask in this week.

Where it goes next: the derivative is a limit of a difference quotient, and that quotient is always 0/0 at substitution, so every derivative from first principles is step 2 of this page. The derivative rules are what you get once that limit has been done in general. L'Hôpital's rule is a later tool for 0/0 that uses derivatives; it is not needed here, and using it before you can factor hides the algebra you are supposed to be learning. Limits at infinity, where x grows without bound, decide the horizontal asymptotes of a rational function and are on the asymptotes page.

Two study notes. The table, the graph and the algebra are three ways of looking at one thing; do all three on a new limit until they agree without effort. And do the sibling before you read its answer: Roediger and Karpicke (2006) found that students who tested themselves on material remembered more a week later than students who reread it, and testing yourself before you reread is the method. Sweller and Cooper (1985) is the reason the worked example comes first.

Limits on one screen

• The definition in words: the value f(x) approaches as x approaches a from both sides; not f(a).
• The three-step order: substitute; if 0/0, factor or rationalise; cancel and substitute again.
• The two algebra moves: factor a polynomial; multiply a root by its conjugate, top and bottom.
• 0/0 is not an answer. It is an instruction.
• The limit laws: sums, differences, products, quotients (bottom's limit not 0), powers and roots pass through; constants come out.
• Continuity: f(a) exists; the limit exists; they are equal.