AP Calculus AB and BC

Limit of (sqrt(x + 9) - 3)/x at 0 Is 1/6

Q: What is the conjugate of \sqrt{x+9}-3?

It is \sqrt{x+9}+3, the same two terms with the sign between them flipped.

Q: Can I use L'Hopital's rule instead?

Yes. The derivative of \sqrt{x+9} is \frac{1}{2\sqrt{x+9}}, and over 1 that is \frac{1}{6} at x = 0. The conjugate method needs no calculus, which is why Unit 1 expects it.

The limit of the square root of x plus 9, minus 3, all over x, as x approaches 0 is one sixth. Multiplying top and bottom by the conjugate turns the numerator into x, which cancels the denominator. What is left is 1 over the square root of x plus 9, plus 3, and that is 1 over 6.

\lim_{x \to 0} \frac{\sqrt{x+9}-3}{x} = \frac{1}{6}

Settled by multiplying by the conjugate.

Multiply by the conjugate

The conjugate of $\sqrt{x+9}-3$ is $\sqrt{x+9}+3$ . Multiplying the pair uses $(a-b)(a+b) = a^{2}-b^{2}$ , and squaring the root is exactly what removes it.

\frac{\sqrt{x+9}-3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} = \frac{(x+9)-9}{x\left(\sqrt{x+9}+3\right)} = \frac{x}{x\left(\sqrt{x+9}+3\right)}

\lim_{x \to 0}\frac{1}{\sqrt{x+9}+3} = \frac{1}{3+3} = \frac{1}{6}

Read the leftover denominator

The numerator collapses to $x$ and nothing else, which is precisely the factor the denominator needs. Leave the new denominator in factored form as $x\left(\sqrt{x+9}+3\right)$ : multiplying it out buries the cancellation you are trying to make.

After the cancellation the surviving expression is continuous at $0$ , so substitution finishes the job. The conjugate does not change the value of anything, it only moves the difficulty from the numerator to a denominator that is safe to evaluate.

The same limit as a derivative

With f(x) = sqrt(x + 9), the quotient is f(x) minus f(0), all over x, so the limit is f prime at 0. Since the derivative is 1 over twice the square root of x plus 9, its value at 0 is 1 over 6.

The mistakes students make

Two of these are algebra with roots, one is a misread of the indeterminate form.

Writing $\sqrt{x+9} = \sqrt{x}+3$ . That collapses the quotient to $\frac{1}{\sqrt{x}}$ , which blows up as $x \to 0^{+}$ and is not even defined for $x < 0$ , so it points nowhere near $\frac{1}{6}$ . Roots do not split across a sum.
Expanding the denominator to $x\sqrt{x+9}+3x$ , which hides the shared factor $x$ and sends you back to $\frac{0}{0}$ .
Answering $0$ because the numerator tends to $0$ . The denominator does too, so the answer depends on which one shrinks faster.

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 $\frac{\sqrt{x+9}-3}{x}$ as x approaches 0?

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

What is the conjugate of $\sqrt{x+9}-3$ ?

It is $\sqrt{x+9}+3$ , the same two terms with the sign between them flipped.

Can I use L'Hopital's rule instead?

Yes. The derivative of $\sqrt{x+9}$ is $\frac{1}{2\sqrt{x+9}}$ , and over $1$ that is $\frac{1}{6}$ at $x = 0$ . The conjugate method needs no calculus, which is why Unit 1 expects it.