AP Calculus AB and BC

Limit of sqrt(x^2+4) - x at Infinity Is 0

The limit of the square root of x squared plus 4, minus x, as x approaches infinity, is 0. Multiplying by the conjugate leaves 4 over the square root of x squared plus 4 plus x, and a constant over an unbounded denominator goes to 0.

lim⁡x→∞x2+4−x=0\lim_{x \to \infty} \sqrt{x^{2}+4}-x = 0

Settled by multiplying by the conjugate.

The conjugate again

(x2+4−x)⋅x2+4+xx2+4+x=4x2+4+x⟶0\left(\sqrt{x^{2}+4}-x\right)\cdot\frac{\sqrt{x^{2}+4}+x}{\sqrt{x^{2}+4}+x} = \frac{4}{\sqrt{x^{2}+4}+x} \longrightarrow 0

Why this one vanishes and a similar one does not

Adding a CONSTANT under the root leaves a constant numerator, which loses to the growing denominator. Adding a term in x, as in sqrt(x^2+x) - x, leaves a numerator that grows too, and the limit is 1/2 instead.

What it says about the graph

The line y=xy = x is a slant asymptote of y=x2+4y = \sqrt{x^{2}+4}, approached from above since the square root is always the larger. The gap closes like 2x\frac{2}{x}.

The mistakes students make

  • Answering 22 from 4\sqrt{4}. The constant under the root does not survive.
  • Answering ∞\infty because both parts are unbounded.

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+4) - x at infinity?

It is 00.

Why does sqrt(x^2+x) - x give 1/2 instead?

Because the conjugate leaves xx on top rather than a constant, and that grows at the same rate as the denominator.