When you try direct substitution and land on \tfrac{0}{0}, don't give up — and don't conclude the limit is zero, or one, or undefined. The form \tfrac{0}{0} is indeterminate: it could be anything. It's a signal that the function has a removable trouble spot you can simplify away.

The classic example, which we first met as 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.

Factor and cancel

Both top and bottom hitting zero at x = 1 is a clue: they share the factor (x - 1). Factor the top, then cancel:

The troublesome factor is gone. Now substitution works on what's left: \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x + 1) = 2. We're allowed to cancel because a limit only cares about x near 1, never x = 1 itself — and everywhere else the two expressions are identical.

The graph: same line, with a hole

Because \tfrac{x^2-1}{x-1} = x + 1 away from x = 1, the graph is the straight line y = x + 1 with a single point missing at x = 1. The cancelling is exactly what "fills in" that hole. Step a point in from both sides and watch it aim at the open circle.

The recipe, and another example

1. Substitute. If you get a real number, you're done.
2. If you get \tfrac{0}{0}, factor top and bottom.
3. Cancel the shared factor, then substitute into what remains.

One more, with a difference of squares on top:

Watch Sal factor a limit