AP Calculus AB and BC

Limit of (x^2-1)/(x-1) as x Approaches 1 Is 2

Q: Why is the limit 2 when f(1) does not exist?

A limit is decided by values near x = 1, never by the value at x = 1. Every input close to 1 produces an output close to 2, and that is the entire test. The gap in the graph sits at a single point, and single points cannot change a limit.

Q: Is this a removable discontinuity?

Yes. Because \lim_{x \to 1} f(x) = 2 exists but f(1) is undefined, redefining f(1) = 2 removes the break and makes the function continuous. Discontinuities that cannot be repaired this way are jumps and infinite discontinuities, where the limit itself fails.

Q: What is the limit of the same function as x \to -1?

It is 0. At x = -1 the denominator is -2, which is not zero, so the function is continuous there and substitution finishes immediately: \frac{(-1)^2-1}{-1-1} = \frac{0}{-2} = 0. Only x = 1 needs the factoring.

The limit of (x^2 - 1)/(x - 1) as x approaches 1 is 2. Direct substitution gives 0/0, so factor the difference of squares: (x - 1)(x + 1) over (x - 1) cancels to x + 1, which is 2 at x = 1. The function has a hole there, so f(1) is undefined while the limit exists.

\lim_{x \to 1} \frac{x^2-1}{x-1} = 2

Settled by factoring and cancelling.

Factoring the difference of squares

The numerator is $x^2 - 1^2$ , a difference of squares, so it splits into two linear factors and one of them is the denominator.

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

For every $x \neq 1$ the common factor is a nonzero number over itself, so it cancels. The cancellation is legitimate precisely because a limit at $1$ never evaluates at $1$ ; it only uses nearby values, where $x - 1 \neq 0$ .

\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} (x+1) = 1 + 1 = 2

What is left, $x + 1$ , is a polynomial and therefore continuous everywhere, so substitution finally works and returns $2$ .

What direct substitution gives

At $x = 1$ the numerator is $1 - 1 = 0$ and the denominator is $1 - 1 = 0$ , so the fraction reports nothing.

\frac{1^2-1}{1-1} = \frac{0}{0}

Two polynomials vanishing at the same input is the fingerprint of a shared factor. The factor theorem says that if a polynomial is zero at $x = 1$ then $(x-1)$ divides it, so the cancellation is guaranteed before you attempt it. That is why $\frac{0}{0}$ on a rational function is an instruction to factor, not a dead end.

The hole is the whole story

The graph of $\frac{x^2-1}{x-1}$ is the line $y = x + 1$ with one point removed at $(1, 2)$ . The limit is the height of that missing point. Defining the value there to be $2$ would patch the function into a continuous one, which is what makes the discontinuity removable.

The mistakes students make

Reporting that the limit does not exist because $f(1)$ is undefined. The limit asks where the outputs are heading, not what happens on arrival.
Reading $\frac{0}{0}$ as $1$ , or as $0$ , or as undefined. It is indeterminate, which means the answer is still open.
Cancelling and then forgetting the restriction. $\frac{x^2-1}{x-1}$ and $x + 1$ agree at every input except $x = 1$ , where the first has no value at all, so they are not the same function.
Reaching for L'Hopital's rule. It does give $\frac{2x}{1} = 2$ , but factoring is quicker and is the skill Unit 1 is checking.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is the limit $2$ when $f(1)$ does not exist?

A limit is decided by values near $x = 1$ , never by the value at $x = 1$ . Every input close to $1$ produces an output close to $2$ , and that is the entire test. The gap in the graph sits at a single point, and single points cannot change a limit.

Is this a removable discontinuity?

Yes. Because $\lim_{x \to 1} f(x) = 2$ exists but $f(1)$ is undefined, redefining $f(1) = 2$ removes the break and makes the function continuous. Discontinuities that cannot be repaired this way are jumps and infinite discontinuities, where the limit itself fails.

What is the limit of the same function as $x \to -1$ ?

It is $0$ . At $x = -1$ the denominator is $-2$ , which is not zero, so the function is continuous there and substitution finishes immediately: $\frac{(-1)^2-1}{-1-1} = \frac{0}{-2} = 0$ . Only $x = 1$ needs the factoring.