AP Calculus AB and BC

Limit of (4x^2-1)/(2x-1) as x Approaches 1/2

Q: Why is the limit point a fraction?

Because the denominator vanishes there: 2x - 1 = 0 at x = \frac{1}{2}. The limit point is always where the trouble is, whatever number that turns out to be.

The limit of four x squared minus one over two x minus one as x approaches one half is two. The numerator is a difference of squares whose first factor is exactly the denominator, so the cancellation is immediate.

\lim_{x \to \frac{1}{2}} \frac{4x^{2}-1}{2x-1} = 2

Settled by factoring a difference of squares.

Spot the difference of squares

4x^{2}-1 = (2x)^{2}-1^{2} = (2x-1)(2x+1)

The first factor is exactly the denominator, so it cancels and leaves $2x+1$ . Substituting $x = \frac{1}{2}$ gives $1 + 1 = 2$ .

The step people miss is recognising $4x^{2}$ as $(2x)^{2}$ rather than $4 \cdot x^{2}$ . Once the coefficient is folded inside the square, the pattern is the standard one.

A useful sanity check

The simplified function is the line $y = 2x+1$ with a hole at $x = \frac{1}{2}$ . Evaluating anywhere nearby, say $x = 0.51$ , gives $2.02$ , close to the answer, which is a quick way to catch an algebra slip.

Testing a nearby value costs seconds and catches sign errors and dropped factors that a re-read of the algebra often will not.

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 point a fraction?

Because the denominator vanishes there: $2x - 1 = 0$ at $x = \frac{1}{2}$ . The limit point is always where the trouble is, whatever number that turns out to be.

How do I spot a difference of squares with coefficients?

Check whether both terms are perfect squares once coefficients are absorbed. $4x^{2}$ is $(2x)^{2}$ , $9x^{2}$ is $(3x)^{2}$ , and $25x^{4}$ is $(5x^{2})^{2}$ .