Multivariable calculus

Limit of (x^2+y^2)/(sqrt(x^2+y^2+1)-1) Is 2

Q: Is there a quicker route than the conjugate?

Yes, if you know that \sqrt{1+u} \approx 1 + \frac{u}{2} for small u. The denominator is then about \frac{u}{2}, and the quotient is about \frac{u}{u/2} = 2. The conjugate multiplication is the same idea carried out exactly instead of approximately.

Q: What is the limit of (sqrt(x^2+y^2+1)-1)/(x^2+y^2)?

It is \frac{1}{2}, the reciprocal of this page's answer. The same conjugate step turns it into \frac{1}{\sqrt{x^2+y^2+1}+1}, which is continuous at the origin with value \frac{1}{2}.

The limit of (x^2 + y^2) / (sqrt(x^2 + y^2 + 1) - 1) as (x, y) approaches the origin is 2. Substitution gives the indeterminate form 0/0. Multiplying top and bottom by the conjugate sqrt(x^2 + y^2 + 1) + 1 simplifies the quotient to exactly that expression, which is continuous at the origin with value 2.

f(x,y) = \frac{x^2+y^2}{\sqrt{x^2+y^2+1} - 1}

The limit exists

\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1} - 1} = 2

Path	Limit along it
along y = 0	2
along y = x	2
along x = 0	2

Substitution gives 0 over 0

At the origin the numerator is $0$ and the denominator is $\sqrt{1} - 1 = 0$ . That is an indeterminate form, which means the direct approach failed, not that the limit failed.

f(t,0) = \frac{t^2}{\sqrt{t^2+1} - 1} \to 2

f(t,t) = \frac{2t^2}{\sqrt{2t^2+1} - 1} \to 2

The paths point at $2$ . As always, they supply a candidate and nothing more, so the next step is to get the value by an argument that covers every approach.

Multiply by the conjugate

A difference of a square root and a constant is cleared by multiplying by the matching sum, which turns the denominator into a difference of squares.

\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1} \cdot \frac{\sqrt{x^2+y^2+1}+1}{\sqrt{x^2+y^2+1}+1} = \frac{\left(x^2+y^2\right)\left(\sqrt{x^2+y^2+1}+1\right)}{\left(x^2+y^2+1\right) - 1}

= \frac{\left(x^2+y^2\right)\left(\sqrt{x^2+y^2+1}+1\right)}{x^2+y^2} = \sqrt{x^2+y^2+1}+1

The final expression is continuous at the origin, so its limit is its value there.

\lim_{(x,y) \to (0,0)} \left(\sqrt{x^2+y^2+1}+1\right) = \sqrt{1}+1 = 2

Setting $u = x^2+y^2$ makes it a one variable statement: $\frac{u}{\sqrt{u+1}-1} = \sqrt{u+1}+1 \to 2$ as $u \to 0^+$ . No direction ever enters, which is why no path could have disagreed.

The mistake: reading 0/0 as a verdict

Indeterminate means undetermined by that particular calculation. A $\frac{0}{0}$ form can end up as any number, as $\pm\infty$ , or with no limit at all, depending entirely on the relative rates. Here both numerator and denominator vanish to first order in $u$ , and the ratio of their leading coefficients is $2$ .

There is a second trap, and it is a numerical one. Plot or tabulate this function very close to the origin and the values drift away from $2$ , because $\sqrt{u+1}-1$ loses almost all of its significant digits to cancellation once $u$ is around $10^{-12}$ . That is a defect of floating point arithmetic, not of the limit.

The conjugate form $\sqrt{u+1}+1$ is the numerically honest way to evaluate the same quantity, and it stays at $2$ to full precision. Rationalising is therefore two useful things at once: the step that proves the limit, and the rewrite that makes the function computable.

Path values are read near t = 1e-4. Closer in than that, sqrt(x^2+y^2+1) - 1 loses most of its significant digits to floating point cancellation, so a double precision evaluation drifts away from 2 even though the exact limit is 2. A 60 digit evaluation at r = 1e-6 returns 2.0000000000005, and the algebraically identical form sqrt(x^2+y^2+1) + 1 is stable at every radius.

Frequently asked questions

Is there a quicker route than the conjugate?

Yes, if you know that $\sqrt{1+u} \approx 1 + \frac{u}{2}$ for small $u$ . The denominator is then about $\frac{u}{2}$ , and the quotient is about $\frac{u}{u/2} = 2$ . The conjugate multiplication is the same idea carried out exactly instead of approximately.

What is the limit of (sqrt(x^2+y^2+1)-1)/(x^2+y^2)?

It is $\frac{1}{2}$ , the reciprocal of this page's answer. The same conjugate step turns it into $\frac{1}{\sqrt{x^2+y^2+1}+1}$ , which is continuous at the origin with value $\frac{1}{2}$ .