AP Calculus AB and BC

Limit of (sqrt(x+4) - 2) / x at 0 Is 1/4

Q: Is there a faster way?

Yes. The expression is the definition of the derivative of \sqrt{x+4} at 0, and that derivative is \frac{1}{2\sqrt{x+4}}, which is \frac{1}{4} there.

The limit of the square root of x plus 4, minus 2, all over x, as x approaches 0 is 1 over 4. Multiplying by the conjugate turns the numerator into x, which cancels the denominator and leaves 1 over the square root of x plus 4 plus 2.

\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{4}

Settled by multiplying by the conjugate.

Clearing the radical

Substitution gives $\frac{0}{0}$ . The conjugate of $\sqrt{x+4} - 2$ is $\sqrt{x+4} + 2$ , and multiplying by it turns the difference of roots into a difference of squares.

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

Cancelling the $x$ leaves something continuous at $0$ , so substitution finishes it.

\lim_{x \to 0}\frac{1}{\sqrt{x+4}+2} = \frac{1}{2+2} = \frac{1}{4}

Leave the conjugate factored

Do not expand the denominator you just created. The whole point is that the numerator simplifies and cancels; multiplying out the bottom destroys the structure you were after.

It is a derivative too

With $f(x) = \sqrt{x+4}$ and $f(0) = 2$ , the quotient is the definition of $f'(0)$ , which gives a one-line check.

f'(x) = \frac{1}{2\sqrt{x+4}} \implies f'(0) = \frac{1}{2 \cdot 2} = \frac{1}{4}

Recognising the derivative shape is faster than the conjugate, and on a multiple choice question it is often the difference between finishing and running out of time.

The mistakes students make

Writing $\sqrt{x+4} = \sqrt{x} + 2$ . Roots do not open across a sum; at $x = 5$ that claims about $4.24$ where the true value is $3$ .
Expanding the conjugate in the denominator, which hides the cancellation.
Cancelling the $x$ before multiplying by the conjugate. There is nothing to cancel until the numerator has been rationalised.

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+4) - 2) / x as x approaches 0?

It is $\frac{1}{4}$ .

Why multiply by the conjugate?

It turns a difference of square roots into a difference of squares, which removes the radical and exposes the factor that cancels.

Is there a faster way?

Yes. The expression is the definition of the derivative of $\sqrt{x+4}$ at $0$ , and that derivative is $\frac{1}{2\sqrt{x+4}}$ , which is $\frac{1}{4}$ there.