A Half, from One Multiplication and No Calculus

Evaluate

Both pieces run off to infinity, so the reflex answer is zero. The answer is $1/2$, it needs one multiplication and no calculus, and the route to it is the same move that handles every difference of two square roots you will ever meet.

What the reflex actually establishes

The argument for zero is that $\sqrt{x^2+x}$ behaves like $x$ for large $x$, so the two terms cancel. The first half is true and the conclusion does not follow. "Behaves like" is a statement about the ratio:

Two quantities can have a ratio tending to 1 and a difference tending to anything at all, including infinity. Here the difference is not even small at moderate $x$: at $x = 100$ the gap is 0.4988, and the two curves are 100 apart in absolute terms while their difference is comfortably measurable.

One multiplication, exactly

Multiply the expression by its conjugate over itself. Nothing is approximated and no series is involved:

The roots cancel because a difference of squares removes them, and the top collapses to a bare $x$. So for every $x \ge 0$,

This is an identity, not an estimate. The difficulty in (1) was that a large quantity was being subtracted from another large quantity, and equation (4) has replaced that with a ratio of two large quantities, which is a shape we know how to read.

Reading off the limit

Divide the top and bottom of (4) by $x$, which is legal for $x > 0$:

The middle expression is continuous in $1/x$at zero, so the limit is just its value there. No L'Hôpital, no Taylor series, no estimation of anything. The whole problem was the original shape.

The gap never reaches a half

Equation (5) says the gap tends to $1/2$. It says nothing about whether it arrives from above or below, and that is worth settling, because a horizontal line drawn on a graph is a claim. Complete the square:

The excess is exactly $1/4$, for every $x$, so $\sqrt{x^2+x} < x + \tfrac12$ always, and the gap is strictly below $1/2$ forever. It also increases the whole way up, so $1/2$ is the least upper bound and is never attained.

Some measured values, computed in sixty digit arithmetic:

The curve is a hyperbola, and that explains the half

Equation (6) looks like a lucky algebraic accident. It is not. Square both sides of $y = \sqrt{x^2+x}$ and complete the square in $x$:

That is a rectangular hyperbola centred at $(-\tfrac12, 0)$, and its asymptotes are the lines $y = \pm\left(x + \tfrac12\right)$. The curve in figure 1 is the upper branch, and the line it is heading for is $y = x + \tfrac12$ exactly.

So the answer to the original question is the vertical offset of an asymptote, and the whole problem was asking how far the hyperbola sits above the diagonal once you are far enough out. A branch of a hyperbola always stays strictly on one side of its asymptote, which is the same statement as equation (6) and the reason the gap never reaches a half.

How fast it approaches

The rate falls out of the same rewrite. Expanding $\sqrt{1 + 1/x}$ gives

so the shortfall from a half is about $1/(8x)$. At $x = 100$ that predicts $0.5 - 0.00125 = 0.49875$, against the true 0.4987562, an error of six parts in a million. The series is also the fastest way to see why the approach is from below: the first correction is negative.

The two heavier routes, and why they are heavier

The conjugate is not the only way in. Both standard alternatives work, and comparing them is a decent illustration of when machinery earns its keep.

L'Hôpital's rule needs a quotient, so substitute $t = 1/x$ first and write the expression as

Both top and bottom now go to zero, so differentiating each gives $\frac{1}{2\sqrt{1+t}} \to \tfrac12$. Correct, and it required manufacturing a quotient out of a difference, which is exactly what equation (4) did for free and without differentiating anything.

The binomial expansion of $\sqrt{1 + 1/x}$ also works, and it gives more than the limit, since it hands you the rate as well. What it costs is a convergence argument: the series is valid for $|1/x| < 1$, which is fine here but is a condition you have to check rather than an identity you can lean on. The conjugate needs no such caveat, because equation (4) holds at every $x \ge 0$ with no remainder term.

The general statement

Nothing depended on the coefficient being 1. For any $b$ and $c$,

because the conjugate trick leaves $bx + c$ on top over roughly $2x$underneath. The constant $c$ never survives, which is a useful thing to know at a glance: only the linear coefficient matters. As a check, $\sqrt{x^2 + 5x + 7} - x \to 5/2$.

Cube roots behave the same way with a different divisor. Since $A^3 - B^3 = (A-B)(A^2+AB+B^2)$, the denominator picks up three terms instead of two and

which gives $1/3$ for cube roots and $1/2$ for our square root, recovering equation (1) as the case $n = 2$.

Where the reasoning needs care

The rewrite in (4) divided by $\sqrt{x^2+x} + x$, which is fine for $x > 0$ and not fine in general. Two things go wrong on the other side.

First, the domain: $x^2 + x \ge 0$ fails on the open interval $(-1, 0)$, so the expression is not defined there at all.

Second, and more instructively, the limit as $x \to -\infty$ is not $-1/2$ or any finite number. For very negative $x$ we have $\sqrt{x^2+x} \approx |x| = -x$, so the expression is close to $-2x$ and diverges to $+\infty$. The conjugate is what detects this: the denominator $\sqrt{x^2+x} + x$ tends to zero rather than to something large, and dividing by it is the step that quietly assumed a sign. Any time you rationalise, check which of the two terms is the bigger one.

Sources and further reading

• The shape of the original problem — Indeterminate form
• The line drawn in figure 2 — Asymptote
• The expansion in equation (9) — Binomial series

The limit was confirmed symbolically and then measured at twelve magnitudes of $x$, from 10 up to $10^{12}$, in sixty digit decimal arithmetic. The rewrite in equation (4) reproduces the gap to better than one part in $10^{40}$ at every magnitude tested, and the monotone increase was confirmed at 20,000 grid points, all of them strictly under a half.