AP Calculus AB and BC

Limit of (sqrt x - 3)/(x - 9) at 9 Is 1/6

Q: Is there a faster way?

Yes. It is the definition of the derivative of \sqrt{x} at x = 9, and that derivative is \frac{1}{2\sqrt{x}} = \frac{1}{6}.

The limit of the square root of x minus 3, over x minus 9, as x approaches 9 is one sixth. The denominator factors as the square root of x minus 3 times the square root of x plus 3, so the shared factor cancels and substitution finishes it.

\lim_{x \to 9} \frac{\sqrt{x}-3}{x-9} = \frac{1}{6}

Settled by factoring a difference of squares.

The denominator is a difference of squares

x - 9 = \left(\sqrt{x}\right)^{2} - 3^{2} = \left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)

\lim_{x \to 9}\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)} = \lim_{x \to 9}\frac{1}{\sqrt{x}+3} = \frac{1}{6}

Multiplying by the conjugate does the same job; factoring is just the shorter route when the radical is already isolated.

It is a derivative too

With $f(x) = \sqrt{x}$ the quotient is the alternate form of $f'(9)$ , and $f'(x) = \frac{1}{2\sqrt{x}}$ gives $\frac{1}{6}$ directly. Recognising the shape is faster than any algebra.

The mistakes students make

Cancelling the $3$ and the $9$ . Neither is a factor.
Reporting no limit because the denominator vanishes. So does the numerator, and they share a factor.

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 (sqrt x - 3)/(x - 9) at 9?

It is $\frac{1}{6}$ .

Is there a faster way?

Yes. It is the definition of the derivative of $\sqrt{x}$ at $x = 9$ , and that derivative is $\frac{1}{2\sqrt{x}} = \frac{1}{6}$ .