AP Calculus AB and BC

Limit of (x - 2) / (sqrt(x+2) - 2) at 2

Q: How does it relate to the flipped version?

\lim_{x\to2}\frac{\sqrt{x+2}-2}{x-2} = \frac{1}{4}, the reciprocal, because the expressions are reciprocals and neither limit is 0.

The limit of x minus 2, over the square root of x plus 2 minus 2, as x approaches 2 is 4. The radical is in the denominator here, so multiplying by the conjugate rationalises the bottom, cancels the x minus 2, and leaves the square root of x plus 2 plus 2.

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

Settled by multiplying by the conjugate of the denominator.

Rationalising the denominator

At $x = 2$ the denominator is $\sqrt{4} - 2 = 0$ and so is the numerator, so the form is $\frac{0}{0}$ . The conjugate belongs to whichever part carries the radical, which here is the bottom.

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

The new denominator is $x - 2$ , which cancels the numerator's factor exactly.

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

The mirror image of the usual problem

Compare this with the more common arrangement, where the radical sits on top. The technique is identical and only the target moves.

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

The two limits are reciprocals, which is a useful check: $4$ and $\frac{1}{4}$ . That relationship holds because one expression is literally the reciprocal of the other and neither limit is $0$ .

The mistakes students make

Multiplying by the conjugate of the NUMERATOR. It does nothing useful; the radical you need to clear is downstairs.
Expanding $(x-2)\left(\sqrt{x+2}+2\right)$ in the numerator, which buries the factor that is about to cancel.
Writing $\sqrt{x+2} - 2 = \sqrt{x}$ . At $x = 7$ that claims about $2.65$ where the true value is $1$ .

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 (x - 2) / (sqrt(x+2) - 2) as x approaches 2?

It is $4$ .

Which conjugate do I use?

The conjugate of whichever part contains the radical. Here it is the denominator, so multiply top and bottom by $\sqrt{x+2}+2$ .

How does it relate to the flipped version?

$\lim_{x\to2}\frac{\sqrt{x+2}-2}{x-2} = \frac{1}{4}$ , the reciprocal, because the expressions are reciprocals and neither limit is $0$ .