Multivariable calculus

Limit of sin(x^2+y^2)/(x^2+y^2) at the Origin

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

It is 0. The same substitution gives \lim_{u \to 0^+} \frac{1 - \cos u}{u}, and since 1 - \cos u behaves like \frac{u^2}{2} for small u, the quotient behaves like \frac{u}{2} and tends to 0.

The limit of sin(x^2 + y^2) / (x^2 + y^2) as (x, y) approaches the origin is 1. The function depends on x and y only through u = x^2 + y^2, and u tends to 0 from above under every approach, so the two variable limit equals the one variable limit of sin(u)/u, which is 1.

f(x,y) = \frac{\sin\left(x^2 + y^2\right)}{x^2 + y^2}

The limit exists

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

Path	Limit along it
along y = 0	1
along y = x	1
along y = 3x	1

Every path returns 1

Start with three lines of different slopes, to see whether the direction of approach changes anything.

f(t,0) = \frac{\sin\left(t^2\right)}{t^2} \to 1

f(t,t) = \frac{\sin\left(2t^2\right)}{2t^2} \to 1

f(t,3t) = \frac{\sin\left(10t^2\right)}{10t^2} \to 1

Each of these is the same one variable fact in disguise, with a different constant multiplying $t^2$ . The constant cancels, which is a hint that the direction never entered the problem in the first place.

Substitute the single variable u = x^2 + y^2

The function depends on the point only through the combination $x^2 + y^2$ . Write $u = x^2 + y^2$ , so $f = \frac{\sin u}{u}$ .

The substitution is legitimate for two reasons: $u$ is continuous, and $u \to 0$ exactly when $(x,y) \to (0,0)$ , with $u > 0$ at every point other than the origin itself. So the two variable limit is the one sided one variable limit.

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

Polar coordinates say the same thing even more plainly. Since $x^2+y^2 = r^2$ , the function is $\frac{\sin\left(r^2\right)}{r^2}$ with no $\theta$ anywhere in it. A function that does not mention the angle cannot depend on the direction of approach.

The mistake: reaching for L'Hopital in two variables

L'Hopital's rule is a theorem about functions of one variable. There is no two variable version, and when the input is a point there is nothing obvious to differentiate with respect to.

Students often differentiate top and bottom with respect to $x$ , getting $\frac{2x\cos\left(x^2+y^2\right)}{2x} = \cos\left(x^2+y^2\right) \to 1$ , and land on the right answer by luck. The step is unjustified, and on a function that does not reduce to a single variable it will produce a confidently wrong number.

The honest route is the substitution above. It converts the problem into a one variable limit, and there L'Hopital, or the standard $\frac{\sin u}{u}$ fact, applies for real.

Frequently asked questions

Is the substitution u = x^2 + y^2 always allowed?

It is allowed when the function genuinely depends on the point only through $u$ , and when $u \to 0$ precisely as $(x,y) \to (0,0)$ . Both hold here. If a stray $x$ or $y$ survives the substitution, as in $\frac{x\sin\left(x^2+y^2\right)}{x^2+y^2}$ , the reduction is incomplete and you need a bound instead.

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

It is $0$ . The same substitution gives $\lim_{u \to 0^+} \frac{1 - \cos u}{u}$ , and since $1 - \cos u$ behaves like $\frac{u^2}{2}$ for small $u$ , the quotient behaves like $\frac{u}{2}$ and tends to $0$ .