AP Calculus AB and BC

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

The limit of the square root of x squared minus x, minus x, as x approaches infinity is minus one half. The form is infinity minus infinity, which is indeterminate, so multiply by the conjugate to turn it into a quotient the degree rules can handle.

lim⁡x→∞x2−x−x=−12\lim_{x \to \infty} \sqrt{x^{2}-x} - x = -\frac{1}{2}

Settled by multiplying by the conjugate.

Turning a difference into a quotient

Both terms run to infinity, and ∞−∞\infty - \infty decides nothing: the answer depends on how fast each part grows. The conjugate converts the difference into a fraction.

(x2−x−x)⋅x2−x+xx2−x+x=(x2−x)−x2x2−x+x=−xx2−x+x\left(\sqrt{x^{2}-x} - x\right)\cdot\frac{\sqrt{x^{2}-x}+x}{\sqrt{x^{2}-x}+x} = \frac{\left(x^{2}-x\right) - x^{2}}{\sqrt{x^{2}-x}+x} = \frac{-x}{\sqrt{x^{2}-x}+x}

Now divide top and bottom by xx, remembering that for positive xx we have x2=x\sqrt{x^{2}} = x.

−xx2−x+x=−11−1x+1⟶−11+1=−12\frac{-x}{\sqrt{x^{2}-x}+x} = \frac{-1}{\sqrt{1-\frac{1}{x}}+1} \longrightarrow \frac{-1}{1+1} = -\frac{1}{2}

Why the answer is finite at all

x2−x\sqrt{x^{2}-x} is very slightly less than xx for large xx, and the gap does not close: it settles at 12\frac{1}{2}. Completing the square shows why, since x2−x=(x−12)2−14x^{2}-x = \left(x-\frac{1}{2}\right)^{2} - \frac{1}{4}, so the root is just under x−12x - \frac{1}{2}.

That also explains the sign. The square root is the smaller quantity, so subtracting xx leaves something negative.

The general shape

The limit of the square root of x squared plus bx, minus x, at infinity is b over 2. With b equal to minus 1 that gives minus one half, matching this page.

The mistakes students make

  • Answering 00 because both parts behave like xx. They do, but the DIFFERENCE settles at a nonzero constant.
  • Answering ∞\infty or calling ∞−∞\infty - \infty equal to 00. It is an indeterminate form, not a value.
  • Writing x2−x=x−x\sqrt{x^{2}-x} = x - \sqrt{x}. Roots do not distribute across subtraction.
  • Getting the sign backwards. The root is smaller than xx, so the difference is negative.

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^2 - x) - x as x approaches infinity?

It is −12-\frac{1}{2}.

Why is infinity minus infinity indeterminate?

Because the answer depends entirely on the relative rates. The same form can give 00, a finite constant, or infinity, depending on the functions involved.

What is the general result?

lim⁡x→∞(x2+bx−x)=b2\lim_{x\to\infty}\left(\sqrt{x^{2}+bx} - x\right) = \frac{b}{2}.