AP Calculus AB and BC

Limit of (sqrt(x+1) - 1)/x as x Approaches 0 Is 1/2

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

Yes. The form is \frac{0}{0}, and differentiating top and bottom gives \frac{1}{2\sqrt{x+1}}, which is \frac{1}{2} at x = 0. Unit 1 wants the conjugate, since L'Hopital arrives in Unit 4, but it is a quick way to check.

Q: What is the limit of (sqrt(x+a) - sqrt(a))/x?

It is \frac{1}{2\sqrt{a}} for any a > 0, by the same conjugate step: the numerator becomes x and the surviving fraction is \frac{1}{\sqrt{x+a}+\sqrt{a}}. Setting a = 1 recovers \frac{1}{2}, and a = 4 gives \frac{1}{4}.

Q: Can I use the binomial approximation instead?

For a quick check, yes: \sqrt{1+x} \approx 1 + \frac{x}{2} for small x, so the numerator is about \frac{x}{2} and the quotient is about \frac{1}{2}. On a free-response answer, show the conjugate work, since the approximation is an estimate rather than a justification.

The limit of (sqrt(x+1) - 1)/x as x approaches 0 is 1/2. Direct substitution gives 0 over 0, and the radical blocks ordinary factoring. Multiplying top and bottom by the conjugate sqrt(x+1) + 1 turns the numerator into x, which cancels and leaves 1 over sqrt(x+1) + 1, equal to 1/2 at x = 0.

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

Settled by multiplying by the conjugate.

Clearing the radical with a conjugate

Nothing cancels while the root is in the way. Multiplying by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ is multiplying by 1, so the function is untouched, but the numerator becomes a difference of squares and the root disappears.

\frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \frac{(x+1)-1}{x\left(\sqrt{x+1}+1\right)} = \frac{x}{x\left(\sqrt{x+1}+1\right)}

Now the $x$ on top is a genuine factor, so it cancels for every $x \neq 0$ , and the limit ignores $x = 0$ itself. What is left is continuous at the origin, so substitution is finally legal.

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

The same fraction, read as a derivative

With $g(x) = \sqrt{x+1}$ and $g(0) = 1$ , the expression is $\frac{g(x)-g(0)}{x-0}$ , the difference quotient for $g$ at 0. So the limit is $g'(0) = \frac{1}{2\sqrt{0+1}} = \frac{1}{2}$ , which is why this exact limit shows up when the derivative of a square root is built from the definition.

Why direct substitution fails

Since $\sqrt{0+1} = 1$ , the numerator is $1 - 1 = 0$ over a denominator of 0.

\frac{\sqrt{0+1}-1}{0} = \frac{0}{0}

The reflex after $\frac{0}{0}$ is to factor, but there is no polynomial on top to factor. A lone radical set against a plain number is the marker for the conjugate instead. Two-sided is fine here: the domain is $x \ge -1$ , so the interval to the left of 0 is inside it.

$x$	$\frac{\sqrt{x+1}-1}{x}$
-0.1	0.513167
0.1	0.488088
0.01	0.498756

The two sides close in on $\frac{1}{2}$ from opposite directions, which is the numerical signature of a limit that exists and a graph with a single missing point.

The mistakes students make

Writing $\sqrt{x+1} = \sqrt{x}+1$ . Roots do not distribute over addition: at $x = 3$ the left side is 2 and the right side is about $2.73$ .
Multiplying only the numerator by the conjugate. Both parts must be multiplied, or the expression is no longer equal to the original.
Using $\sqrt{x+1}-1$ as its own conjugate. The conjugate flips the sign between the terms, giving $\sqrt{x+1}+1$ .
Cancelling the two 1s to leave $\frac{\sqrt{x}}{x}$ . Terms inside a sum cannot be cancelled across a fraction bar.
Reporting 2 instead of $\frac{1}{2}$ . After the cancellation the surviving expression is $\frac{1}{\sqrt{x+1}+1}$ , so the 2 ends up underneath.

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

Does L'Hopital's rule work here?

Yes. The form is $\frac{0}{0}$ , and differentiating top and bottom gives $\frac{1}{2\sqrt{x+1}}$ , which is $\frac{1}{2}$ at $x = 0$ . Unit 1 wants the conjugate, since L'Hopital arrives in Unit 4, but it is a quick way to check.

What is the limit of (sqrt(x+a) - sqrt(a))/x?

It is $\frac{1}{2\sqrt{a}}$ for any $a > 0$ , by the same conjugate step: the numerator becomes $x$ and the surviving fraction is $\frac{1}{\sqrt{x+a}+\sqrt{a}}$ . Setting $a = 1$ recovers $\frac{1}{2}$ , and $a = 4$ gives $\frac{1}{4}$ .

Can I use the binomial approximation instead?

For a quick check, yes: $\sqrt{1+x} \approx 1 + \frac{x}{2}$ for small $x$ , so the numerator is about $\frac{x}{2}$ and the quotient is about $\frac{1}{2}$ . On a free-response answer, show the conjugate work, since the approximation is an estimate rather than a justification.