AP Calculus AB and BC

Limit of (sqrt x - 2)/(x-4) at x = 4 Is 1/4

Q: Is there a derivative shortcut for this one?

Yes. With g(x) = \sqrt{x} and g(4) = 2, the fraction is \frac{g(x)-g(4)}{x-4}, the difference quotient for \sqrt{x} at x = 4. So the limit is g'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}, matching the conjugate work.

Q: Does L'Hopital's rule work here?

It does. The form is \frac{0}{0}, and differentiating top and bottom gives \frac{1}{2\sqrt{x}}, which is \frac{1}{4} at x = 4. Unit 1 expects the conjugate, since L'Hopital is a Unit 4 tool, but it is a fair way to check the answer.

Q: When do I use the conjugate instead of factoring?

Use it when substitution gives \frac{0}{0} and a square root is standing in the way, on either the top or the bottom. Factoring handles polynomials; the conjugate handles radicals by squaring them away. If the radical is a sum such as \sqrt{x}+\sqrt{3}, the conjugate is \sqrt{x}-\sqrt{3}.

The limit of (sqrt(x) - 2)/(x - 4) as x approaches 4 is 1/4. Direct substitution gives 0/0, and the radical blocks ordinary factoring. Multiply top and bottom by the conjugate sqrt(x) + 2: the numerator becomes x - 4, cancels, and leaves 1/(sqrt(x) + 2), which is 1/4 at x = 4.

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

Settled by multiplying by the conjugate.

Clearing the radical with a conjugate

The blocker is the radical: $\sqrt{x}-2$ has no polynomial factor to match the $x-4$ underneath it. Multiplying top and bottom by the conjugate $\sqrt{x}+2$ turns the numerator into a difference of squares and clears the root.

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

Multiplying by $\frac{\sqrt{x}+2}{\sqrt{x}+2}$ is multiplying by $1$ , so nothing about the function changed. What changed is that $(x-4)$ is now visible on top, and it cancels for every $x \neq 4$ .

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

The same move, seen as factoring

For $x \ge 0$ the denominator itself factors: $x - 4 = \left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)$ . Cancelling directly leaves $\frac{1}{\sqrt{x}+2}$ in one step. The conjugate is the mechanical version of that observation, which is why it is the standing response to a radical inside $\frac{0}{0}$ .

What direct substitution gives

Since $\sqrt{4} = 2$ , the numerator is $2 - 2 = 0$ , and the denominator is $4 - 4 = 0$ .

\frac{\sqrt{4}-2}{4-4} = \frac{0}{0}

On a rational function the reflex after $\frac{0}{0}$ is to factor, but there is no polynomial on top to factor. A square root inside an indeterminate quotient is the marker for the conjugate instead, and the tell is that the radical sits alone against a plain number.

Nothing here can be cancelled before that step. Writing $\frac{\sqrt{x}}{x}$ and reducing it, or cancelling the $2$ with the $4$ , changes the function into a different one.

The mistakes students make

Multiplying only the numerator by $\sqrt{x}+2$ . Both parts must be multiplied, or the expression is no longer equal to the original.
Using $\sqrt{x}-2$ as its own conjugate. The conjugate flips the sign between the terms, so it is $\sqrt{x}+2$ .
Writing $\sqrt{x}-2 = \sqrt{x-4}$ . Roots do not distribute across subtraction; at $x = 8$ the first is about $0.83$ and the second is $2$ .
Reporting $4$ instead of $\frac{1}{4}$ . After the cancellation the surviving expression is $\frac{1}{\sqrt{x}+2}$ , so the $4$ ends up in the denominator.

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

Is there a derivative shortcut for this one?

Yes. With $g(x) = \sqrt{x}$ and $g(4) = 2$ , the fraction is $\frac{g(x)-g(4)}{x-4}$ , the difference quotient for $\sqrt{x}$ at $x = 4$ . So the limit is $g'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}$ , matching the conjugate work.

Does L'Hopital's rule work here?

It does. The form is $\frac{0}{0}$ , and differentiating top and bottom gives $\frac{1}{2\sqrt{x}}$ , which is $\frac{1}{4}$ at $x = 4$ . Unit 1 expects the conjugate, since L'Hopital is a Unit 4 tool, but it is a fair way to check the answer.

When do I use the conjugate instead of factoring?

Use it when substitution gives $\frac{0}{0}$ and a square root is standing in the way, on either the top or the bottom. Factoring handles polynomials; the conjugate handles radicals by squaring them away. If the radical is a sum such as $\sqrt{x}+\sqrt{3}$ , the conjugate is $\sqrt{x}-\sqrt{3}$ .