Limits of the Form 0/0

You try direct substitution, and the calculator of your mind returns \tfrac{0}{0}. Most people's instinct is that something has gone wrong — the limit must be zero, or one, or "undefined", or the question must be broken. None of those. \tfrac{0}{0} is not an answer at all. It's a coded message, and once you learn to read it, it says something very specific:

"The top and bottom share a hidden common factor. Find it, cancel it, and ask again."

A fraction only lands on \tfrac{0}{0} when the numerator and the denominator both vanish at the same point — and two polynomials that both vanish at x = c must both contain the factor (x - c). That shared factor is the culprit: it forces the \tfrac{0}{0}, and it hides the true limit behind it. Your job on this page is to become a code-breaker — factor, cancel, and read off the real answer.

And this is not a niche trick. It is arguably the most important indeterminate form in all of calculus, because — as you'll see near the end of this page — every derivative you will ever compute is secretly a \tfrac{0}{0} limit. Master this move and you've built the engine room of the whole subject.

The classic: factor and cancel

Start with the example we first met in the idea of a limit: \lim_{x \to 1} \frac{x^2 - 1}{x - 1}. Plug in x = 1: \tfrac{1 - 1}{1 - 1} = \tfrac{0}{0}. Indeterminate — the coded message. Decode it: both top and bottom hit zero at x = 1, so both must contain the factor (x - 1). The bottom wears it openly; the top hides it inside a difference of squares. Factor the top, then cancel:

\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad (x \ne 1).

The troublesome factor is gone, and substitution now works on what's left: \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2.

Why is the cancellation legal? Because a limit only cares about x near 1 — never x = 1 itself. At every point except x = 1, the fraction \tfrac{x^2-1}{x-1} and the line x + 1 are literally the same number, so they must be heading toward the same place. Cancelling never touches the one point the limit ignores anyway.

The graph: same line, with a hole

The algebra has a picture, and it's a strangely satisfying one. Because \tfrac{x^2-1}{x-1} = x + 1 away from x = 1, the graph of the original fraction is the straight line y = x + 1 with a single point punched out at x = 1 — an open circle floating where (1, 2) should be. The function has no value there, but the limit doesn't need one: it only asks where the nearby points are aiming.

Step the point in from either side with the slider below and watch its coordinates. From the left it reads f(0.9) = 1.9, f(0.95) = 1.95; from the right f(1.1) = 2.1, f(1.05) = 2.05. Both sides converge on the open circle at height 2. Cancelling the common factor is exactly the algebra that "fills in" that hole — which is why this kind of trouble spot is called removable.

Here is the trap that gives "indeterminate" its name. Students sometimes learn a fake rule — "\tfrac{0}{0} means the answer is 1, because anything over itself is one" or "the zeros cancel, so it's 0". Both are wrong, and here's the proof. These two limits both give \tfrac{0}{0} at x = 5:

\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = 10, \qquad \lim_{x \to 5} \frac{3x - 15}{x - 5} = \lim_{x \to 5} \frac{3(x-5)}{x-5} = 3.

Same coded message, completely different decoded answers — one heads to 10, the other to 3. In fact, for any number k you like, \tfrac{k(x-5)}{x-5} gives \tfrac{0}{0} at x=5 and has limit k. The form \tfrac{0}{0} can conceal any value whatsoever — that is precisely why it's called indeterminate: the form alone determines nothing. Only the algebra behind it does.

Two more traps while we're here:

Never "cancel the zeros". You cancel factors, not values. The move \tfrac{(x-1)(x+1)}{x-1} = x+1 cancels the algebraic factor (x-1), which is a perfectly good nonzero number at every x \ne 1. Writing \tfrac{0}{0} = 1 and crossing out the zeros is division by zero, and it will happily "prove" nonsense.
\tfrac{k}{0} with k \ne 0 is NOT indeterminate. If substitution gives a nonzero top over a zero bottom — say \tfrac{7}{0} — there is no hidden factor and no code to crack: the fraction genuinely blows up near the point (its size grows without bound), so no finite limit exists. Don't waste time factoring; that limit fails for real. The indeterminate case is exclusively zero over zero, where top and bottom race each other down and the winner is anyone's guess until you factor.

The recipe

Everything on this page follows one three-step script. It's worth memorising, because it's the script for a huge fraction of the limit questions you'll ever meet:

Substitute. If you get an honest real number, you're done — that's the limit.
If you get \tfrac{0}{0}, factor top and bottom. The point x = c you're approaching tells you exactly which factor to hunt for: (x - c) must divide both.
Cancel the shared factor, then substitute into what remains.

Here it is on a difference of squares, start to finish. Substituting x = 3 into \tfrac{x^2-9}{x-3} gives \tfrac{0}{0}, so we hunt for the factor (x - 3):

\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x + 3) = 6.

And the picture is the same story as before: the graph below is the line y = x + 3 that the fraction secretly equals — with one invisible pinprick at x = 3, where the original expression has no value but the limit reads 6 off the surrounding points.

Worked example 1: x \to 5

Evaluate \lim_{x \to 5} \frac{x^2 - 25}{x - 5}.

Step 1 — substitute. Top: 5^2 - 25 = 0. Bottom: 5 - 5 = 0. The form is \tfrac{0}{0} — indeterminate, so the message says: find the common factor (x - 5).

Step 2 — factor. The top is a difference of squares, x^2 - 25 = (x-5)(x+5), and there's the hidden factor:

\frac{x^2 - 25}{x - 5} = \frac{(x-5)(x+5)}{x-5} = x + 5 \quad (x \ne 5).

Step 3 — cancel and substitute.

\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} (x + 5) = 10.

Notice the pattern quietly emerging: \tfrac{x^2 - c^2}{x - c} \to 2c as x \to c, because after cancelling you're left with x + c, which heads to c + c. At c = 1, 3, 5 we've now seen the answers 2, 6, 10 — always double the point of approach.

Worked example 2: quadratic over quadratic

The factor-hunt works just as well when both top and bottom need factoring. Evaluate \lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4}.

Step 1 — substitute. Top: 4 + 2 - 6 = 0. Bottom: 4 - 4 = 0. So \tfrac{0}{0}, and the shared factor must be (x - 2).

Step 2 — factor both. The top is a quadratic whose factors multiply to -6 and sum to +1: that's +3 and -2. The bottom is another difference of squares:

\frac{x^2 + x - 6}{x^2 - 4} = \frac{(x + 3)(x - 2)}{(x - 2)(x + 2)} = \frac{x + 3}{x + 2} \quad (x \ne 2).

Step 3 — cancel and substitute. With the (x-2) gone, the remaining fraction is perfectly well-behaved at x = 2:

\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4} = \frac{2 + 3}{2 + 2} = \frac{5}{4}.

A limit that is neither 0 nor a whole number — \tfrac{5}{4} — from a form that looked like \tfrac{0}{0}. One more piece of evidence that the coded message can hide absolutely anything.

Worked example 3: expand first — and a glimpse of the future

Sometimes the common factor isn't waiting to be factored out — it only appears after you expand. Evaluate \lim_{h \to 0} \frac{(2+h)^2 - 4}{h}.

Step 1 — substitute. Top: (2+0)^2 - 4 = 0. Bottom: 0. Indeterminate again — but the top doesn't factor nicely as it stands. So expand it:

\frac{(2+h)^2 - 4}{h} = \frac{4 + 4h + h^2 - 4}{h} = \frac{4h + h^2}{h}.

Step 2 — now the factor shows itself. The 4s cancelled, and every surviving term in the numerator carries an h:

\frac{4h + h^2}{h} = \frac{h\,(4 + h)}{h} = 4 + h \quad (h \ne 0).

Step 3 — substitute.

\lim_{h \to 0} \frac{(2+h)^2 - 4}{h} = \lim_{h \to 0} (4 + h) = 4.

Now look hard at that limit, because you will meet it again — this is exactly the shape of a derivative. With f(x) = x^2, the expression \tfrac{(2+h)^2 - 4}{h} is \tfrac{f(2+h) - f(2)}{h}: the average slope of the parabola over a small step h, and its limit as h \to 0 is the slope of the curve at x = 2. That idea gets a page of its own — the derivative at a point — but the engine driving it is the one you just used: expand, cancel the h, substitute.

Here's the reveal promised at the top. The derivative — the central object of differential calculus, the thing that gives you instantaneous speed, tangent slopes, rates of change of anything — is defined as f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Now substitute h = 0 into that fraction, for any function f and any point a: the top reads f(a) - f(a) = 0, the bottom reads 0. Every single derivative, of every function, at every point, is a \tfrac{0}{0} limit in disguise. There are no exceptions.

That's why this form is the one worth mastering. When you compute f'(3) next term, or the velocity of a rocket from its height function, you'll be doing precisely what you did on this page: staring at a \tfrac{0}{0}, refusing to panic, and algebraically dissolving the shared factor until the true value steps out from behind the code. Calculus doesn't avoid dividing zero by zero — it builds a machine for sneaking up on it, and this page is that machine.

Much later in the story there's even a power tool for stubborn \tfrac{0}{0}s that refuse to factor: L'Hôpital's rule, which — deliciously — uses derivatives to crack the very form that derivatives are made of. Its history is a scandal in its own right: the wealthy Marquis de l'Hôpital paid the brilliant, cash-strapped Johann Bernoulli a retainer for his mathematical discoveries, published the rule in his own 1696 textbook, and has had his name on Bernoulli's theorem ever since. Money can, apparently, buy immortality.

Polynomials always surrender their (x - c) factor eventually — but what about this one? \lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}. Substituting x = 0 gives \tfrac{\sqrt{1} - 1}{0} = \tfrac{0}{0}, but a square root isn't a polynomial: there's no visible factor to cancel. The trick is to manufacture one by multiplying top and bottom by the conjugate, \sqrt{x+1} + 1 — the same expression with the sign flipped. Because (A - B)(A + B) = A^2 - B^2, the roots on top square themselves away:

\frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \frac{(x + 1) - 1}{x\left(\sqrt{x+1} + 1\right)} = \frac{x}{x\left(\sqrt{x+1} + 1\right)} = \frac{1}{\sqrt{x+1} + 1}.

There was the hidden factor of x all along — the conjugate just flushed it into the open. Now substitute:

\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}.

The graph confirms it: a smooth curve with a single hole at x = 0, hovering at height exactly \tfrac{1}{2}.

Watch Sal factor a limit

Sal Khan works a factor-and-cancel limit from scratch — the same recipe as above, narrated step by step. Watch for the moment he pauses over the x \ne c caveat: that's the "a limit never looks at the point itself" idea doing its quiet work.