AP Calculus AB and BC

Limit of sqrt(x^2+x) - x at Infinity

Q: Why can't I use L'Hopital's rule directly?

It applies to quotients in \frac{0}{0} or \frac{\infty}{\infty} form, and \sqrt{x^{2}+x}-x is a difference, not a quotient. The \infty - \infty form has to be rewritten first, and the conjugate is the cleanest rewrite. Once it is \frac{x}{\sqrt{x^{2}+x}+x} the rule becomes legal, though dividing by x is faster.

Q: What is the limit as x approaches negative infinity?

It is \infty, not \frac{1}{2}, and the conjugate is not the way to see it. For x < -1 the radical is roughly |x| = -x, so the expression behaves like -x - x = -2x, which is large and positive. At x = -100 the value is about 199.5. Never carry an at-infinity answer over to the other end without rechecking signs.

Q: Does the conjugate trick always work on infinity minus infinity?

It is the right first move whenever at least one term is a square root, because squaring is what cancels the unbounded parts. Without a radical, use a different rewrite: factor out the dominant term, or put the two pieces over a common denominator, then evaluate the quotient that results.

The limit of the square root of x squared plus x, minus x, as x approaches infinity is one half. Substitution gives infinity minus infinity, an indeterminate form, so multiply by the conjugate: that turns the difference into x over the sum of that root and x, which tends to one half. Answering 0 is the classic error.

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

Settled by multiplying by the conjugate.

Running the conjugate

Multiply and divide by the conjugate $\sqrt{x^{2}+x}+x$ . A difference times its conjugate is a difference of squares, and squaring the radical is exactly what clears it.

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

The $x^{2}$ terms cancel, which is the point of the whole maneuver. A subtraction of two unbounded quantities has become a quotient.

\sqrt{x^{2}+x}-x = \frac{x}{\sqrt{x^{2}+x}+x}

That quotient is $\frac{\infty}{\infty}$ , so divide top and bottom by $x$ . Since $x > 0$ on this side, $x = \sqrt{x^{2}}$ and the division moves cleanly under the radical.

\frac{x}{\sqrt{x^{2}+x}+x} = \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \quad (x > 0)

Now $\frac{1}{x}$ goes to 0, the radical goes to 1, and the denominator settles at 2.

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

$x$	$\sqrt{x^{2}+x}-x$
$10$	$0.48809$
$100$	$0.49876$
$1000$	$0.49988$

Why substitution fails

Both terms grow without bound, so substitution returns $\infty - \infty$ . That form is indeterminate, and it is the one students most often treat as though it were arithmetic. Two unbounded quantities can differ by anything at all, because the answer depends on how the gap between them behaves, not on the fact that both are large.

Four expressions share the form and land in four different places, including two that differ only in a single coefficient under the radical.

Expression	Form	Limit at infinity
$(x+3)-x$	$\infty - \infty$	$3$
$\sqrt{x^{2}+1}-x$	$\infty - \infty$	$0$
$\sqrt{x^{2}+x}-x$	$\infty - \infty$	$\frac{1}{2}$
$x^{2}-x$	$\infty - \infty$	$\infty$

Rows two and three are the instructive pair. Adding a constant under the radical leaves a gap that closes to 0, while adding an $x$ leaves a gap that settles at $\frac{1}{2}$ , and no inspection of the form distinguishes them.

Why 0 is the wrong instinct

The reasoning behind answering 0 is that $\sqrt{x^{2}+x}$ is basically $x$ for large inputs, so the difference should vanish. The first half of that is true as a leading-order statement. The trouble is that the entire question is the size of the leftover, and rounding the radical to $x$ throws the leftover away.

Completing the square shows what the leftover actually is, and confirms the answer without any conjugate.

x^{2}+x = \left(x+\frac{1}{2}\right)^{2} - \frac{1}{4}

So $\sqrt{x^{2}+x}$ sits just under $x + \frac{1}{2}$ , and the deficit shrinks as $x$ grows. Subtracting $x$ leaves something just under $\frac{1}{2}$ , closing on it from below, which matches the table above.

The general version is worth carrying. Whatever sits under the radical, the linear coefficient is what survives, halved.

\lim_{x \to \infty} \left(\sqrt{x^{2}+bx+c}-x\right) = \frac{b}{2}

Check it against the cases you know

Setting $b = 1$ returns $\frac{1}{2}$ , this page's answer. Setting $b = 0$ returns 0, which is the $\sqrt{x^{2}+1}-x$ row of the table, and shows the constant $c$ never affects the limit. The same formula says $y = x + \frac{b}{2}$ is the slant asymptote of $y = \sqrt{x^{2}+bx+c}$ on the right; on the left the radical behaves like $-x - \frac{b}{2}$ , so that branch has the asymptote $y = -x - \frac{b}{2}$ .

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

Why can't I use L'Hopital's rule directly?

It applies to quotients in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, and $\sqrt{x^{2}+x}-x$ is a difference, not a quotient. The $\infty - \infty$ form has to be rewritten first, and the conjugate is the cleanest rewrite. Once it is $\frac{x}{\sqrt{x^{2}+x}+x}$ the rule becomes legal, though dividing by $x$ is faster.

What is the limit as x approaches negative infinity?

It is $\infty$ , not $\frac{1}{2}$ , and the conjugate is not the way to see it. For $x < -1$ the radical is roughly $|x| = -x$ , so the expression behaves like $-x - x = -2x$ , which is large and positive. At $x = -100$ the value is about $199.5$ . Never carry an at-infinity answer over to the other end without rechecking signs.

Does the conjugate trick always work on infinity minus infinity?

It is the right first move whenever at least one term is a square root, because squaring is what cancels the unbounded parts. Without a radical, use a different rewrite: factor out the dominant term, or put the two pieces over a common denominator, then evaluate the quotient that results.