AP Calculus AB and BC

Limit of sqrt(x+1) - sqrt(x) at Infinity Is 0

The limit of the square root of x plus 1 minus the square root of x, as x approaches infinity, is 0. The form is infinity minus infinity, so multiply by the conjugate: the numerator collapses to 1 while the denominator grows without bound.

lim⁡x→∞x+1−x=0\lim_{x \to \infty} \sqrt{x+1}-\sqrt{x} = 0

Settled by multiplying by the conjugate.

The conjugate turns it into a quotient

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

The numerator becomes the constant 11 and the denominator grows without bound, so the limit is 00.

Two square roots grow closer together

The gap between consecutive square roots shrinks like 1/(2 sqrt x). At x = 10000 the difference is about 0.005. Both parts run to infinity, and yet their difference vanishes.

Compare with the version that does not vanish

Change the inside from x+1x+1 to x2+xx^{2}+x and the same conjugate trick gives 12\frac{1}{2} instead of 00. Whether the difference vanishes depends on how the two growth rates compare, not on the fact that both are unbounded.

The mistakes students make

  • Answering ∞\infty because both terms are unbounded. ∞−∞\infty - \infty is indeterminate.
  • Writing x+1=x+1\sqrt{x+1} = \sqrt{x}+1. At x=9x = 9 that claims 44 where the true value is about 3.1623.162.

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+1) - sqrt(x) at infinity?

It is 00.

Why is infinity minus infinity indeterminate?

Because the answer depends entirely on relative rates. The same form gives 12\frac{1}{2} for x2+x−x\sqrt{x^{2}+x} - x.