Limits by factoring

Zero over zero

What is \displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}? Substituting x=2 gives \tfrac{0}{0}, which is not a number. Does that mean there is no limit?

Not necessarily. A limit asks what happens as x gets close to 2, never what happens at 2 itself. Move the ochre point toward the gap in the graph and watch the numerator, the denominator, and their ratio.

Drag the ochre point along the curve, or use the controls. Turn on "Zoom in" to follow the point as it gets very close to a.

Limit

Distance from a

Left of a Right of a

Show the simplified function Zoom in

Limit

Numerator

Denominator

Ratio f(x)

Values of f(x) near a, from both sides

Distance from a	f(a − distance)	f(a + distance)

Why factoring works

If a polynomial is 0 at x=a, then x-a is one of its factors. This is the factor theorem. So when the numerator and the denominator are both 0 at a, they share the factor x-a, and that shared factor is what produces \tfrac00.

\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\qquad\text{for } x\ne 2

Cancelling is allowed only because x\ne 2. The two functions \tfrac{x^2-4}{x-2} and x+2 agree at every x except 2, where the first one is undefined. Their graphs are the same line, except that the first has a hole.

A limit as x\to 2 uses only values with x\ne 2. So the two functions have the same limit, and x+2 is continuous, so its limit is found by substituting: 2+2=4. Turn on "Show the simplified function" in the graph. Where do the two graphs differ?

Solve it step by step

Each problem starts the same way: substitute and look at what you get. Some problems need factoring and some do not. Answer each step before moving on. The score counts steps you answer correctly on the first try.

Problems solved0

First-try steps0 of 0

Common mistakes

Writing \tfrac00=1 or \tfrac00=0: \tfrac00 is not a value. It says only that more work is needed. The limit could be any number, or it might not exist.
Treating \tfrac{k}{0} like \tfrac00: If the numerator approaches a nonzero number and the denominator approaches 0, there is no common factor to cancel. The ratio grows without bound, so there is no finite limit. What happens to the sign on each side of a?
Stopping before factoring completely: If you cancel once and still get \tfrac00, there is another common factor. Factor until no factor of x-a remains in the denominator.
Dropping the limit notation: Write \lim_{x\to a} on every line until you substitute. The expressions are equal only for x\ne a, and the limit is what makes that enough.