AP Calculus AB and BC

Limit of (x - 4)/(sqrt x - 2) at 4 Is 4

Q: Do I have to use the conjugate when the root is in the denominator?

No. Factoring x-4 as \left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right) cancels the denominator directly and saves a step.

The limit of x minus 4, over the square root of x minus 2, as x approaches 4 is 4. Writing x minus 4 as the square root of x minus 2 times the square root of x plus 2 cancels the denominator outright. What remains is the square root of x plus 2, which is 4 at x equals 4.

\lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} = 4

Settled by factoring the numerator as a difference of squares in the root.

Factor the numerator in terms of the root

The root sits in the denominator this time, so the useful move is to rewrite the numerator to match it. For $x > 0$ we have $x = \left(\sqrt{x}\right)^{2}$ and $4 = 2^{2}$ , which makes $x-4$ a difference of squares in $\sqrt{x}$ .

x-4 = \left(\sqrt{x}\right)^{2}-2^{2} = \left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)

\lim_{x \to 4}\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2} = \lim_{x \to 4}\left(\sqrt{x}+2\right) = 2+2 = 4

The conjugate route lands in the same place

If the factorisation does not come to mind, multiply top and bottom by $\sqrt{x}+2$ , the conjugate of the denominator.

\frac{x-4}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)\left(\sqrt{x}+2\right)}{x-4} = \sqrt{x}+2

Both routes end at $\sqrt{x}+2$ . Factoring is one step shorter and it shows straight away why the answer is $4$ rather than something involving a fraction.

The mistakes students make

Each of these produces a clean looking number that is not 4.

Squaring the denominator term by term, writing $\left(\sqrt{x}-2\right)^{2}$ as $x-4$ , so the quotient appears to cancel to $1$ . Squaring a difference is not squaring each term, since $\left(\sqrt{x}-2\right)^{2} = x-4\sqrt{x}+4$ , and squaring the denominator alone changes the quotient anyway.
Reading $\sqrt{x}-2$ as $\sqrt{x-2}$ , which is continuous at $4$ and gives $0$ .
Cancelling the $4$ against the $2$ to leave $\frac{x}{\sqrt{x}}$ and answering $2$ .

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

What is the limit of $\frac{x-4}{\sqrt{x}-2}$ at x = 4?

It is $4$ .

Do I have to use the conjugate when the root is in the denominator?

No. Factoring $x-4$ as $\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)$ cancels the denominator directly and saves a step.

Why is the answer $4$ and not $\frac{1}{4}$ ?

Because the shared factor leaves the denominator, not the numerator. What survives is $\sqrt{x}+2$ , which is $4$ at $x = 4$ . The value $\frac{1}{4}$ belongs to the flipped quotient $\frac{\sqrt{x}-2}{x-4}$ .