Bounding Coordinate Expressions with Norms

Why Bound Coordinates by Distance?

To prove a limit at the origin, it is often useful to estimate a complicated expression using a simpler quantity that depends only on how far $(x,y)$ is from $(0,0)$. The natural distance is $r=\sqrt{x^2+y^2}$. As $(x,y)$ approaches the origin, $r$ approaches zero, and each coordinate is no larger in magnitude than $r$.

That basic observation leads to bounds for products, powers, and polynomials in $x$ and $y$. Once an expression is bounded by a power of $r$, the estimate is independent of the direction of approach. The Squeeze Theorem for multivariable limits then turns a bound tending to zero into a limit proof.

Definition: For a point $(x,y)$, let $r=\sqrt{x^2+y^2}$ denote its distance from the origin. In estimates near the origin, $r>0$ whenever $(x,y)\ne(0,0)$, and $r\to0$ as $(x,y)\to(0,0)$.

The Basic Coordinate Inequalities

Since $x^2$ and $y^2$ are nonnegative, neither can exceed their sum. Thus $x^2\le x^2+y^2$ and $y^2\le x^2+y^2$. Taking nonnegative square roots gives the basic coordinate bounds.

|x|\le \sqrt{x^2+y^2}=r, \qquad |y|\le \sqrt{x^2+y^2}=r.

These inequalities hold at every point in the plane. They imply that a coordinate power is bounded by the corresponding power of $r$. For nonnegative integers $m,n$, the product estimate follows by multiplying the coordinate bounds.

Theorem: For nonnegative integers $m,n$, $$ |x|^m|y|^n\le r^{m+n}. $$ In particular, every monomial of total degree $m+n$ is bounded in absolute value by the same power of the distance.

To see the product estimate directly, raise $|x|\le r$ to the power $m$ and $|y|\le r$ to the power $n$. Both sides are nonnegative, so raising to a nonnegative integer power preserves the inequalities. Multiplying them gives $|x|^m|y|^n\le r^m r^n=r^{m+n}$. If an exponent is zero, the corresponding factor is $1$, and the same conclusion holds.

A sharper estimate is available for a product of one power of each coordinate. Since $(|x|-|y|)^2\ge0$, expanding and rearranging gives $2|xy|\le x^2+y^2$. Therefore:

|xy|\le \frac{x^2+y^2}{2}=\frac{r^2}{2}.

The power bound $|xy|\le r^2$ is already enough for many limits; the factor $1/2$ is useful when a tighter estimate is wanted. The sharper inequality is exact when $|x|=|y|$.

From Coordinate Bounds to Limit Proofs

Suppose a function $f$ is defined near the origin, except possibly at the origin itself. If we can show that $|f(x,y)|$ is at most a function of $r$ that tends to zero as $r\to0$, then $f(x,y)$ must also tend to zero. This is an application of the Squeeze Theorem: the absolute value is nonnegative, and a shrinking upper bound forces it to zero.

Limit-estimate principle: Suppose that for all sufficiently small $r>0$, $$ |f(x,y)|\le C r^p $$ for constants $C>0$ and $p>0$. Since $r\to0$ implies $Cr^p\to0$, the Squeeze Theorem gives $$ \lim_{(x,y)\to(0,0)} f(x,y)=0. $$ The estimate must hold for all sufficiently close points in the domain, not just along selected paths.

The principle can also be expressed in epsilon-delta language. Given $\varepsilon>0$, choose $\delta=(\varepsilon/C)^{1/p}$. If $0<r<\delta$, then $|f(x,y)|\le Cr^p<C\delta^p=\varepsilon$. This verifies the required limit condition whenever the stated bound is valid.

Worked Example: A Mixed Monomial over the Squared Distance

Find the limit of $f(x,y)=\dfrac{x^2y}{x^2+y^2}$ as $(x,y)\to(0,0)$, where the function is defined for $(x,y)\ne(0,0)$.

Write the denominator as $r^2$. The coordinate bounds give $|x|\le r$ and $|y|\le r$, so $|x^2y|=|x|^2|y|\le r^2r=r^3$. For every nonzero point, $r^2>0$, and therefore

\left|\frac{x^2y}{x^2+y^2}\right| = \frac{|x^2y|}{r^2} \le \frac{r^3}{r^2} =r.

As $r\to0$, the upper bound $r$ tends to zero. Also, $0\le |f(x,y)|$, so $0\le |f(x,y)|\le r$ squeezes $|f(x,y)|$ to zero. Hence the limit of $f$ is zero.

For an explicit check, let $\varepsilon>0$ and choose $\delta=\varepsilon$. Whenever $0<\sqrt{x^2+y^2}<\delta$, the estimate gives $|f(x,y)|\le r<\delta=\varepsilon$. Substituting this bound into the limit condition verifies the result.

Bounding Sums of Coordinate Powers

A sum of monomials can be bounded by estimating each term and then adding the bounds. For example, if $P(x,y)=a x^2y+b y^3$, where $a$ and $b$ are constants, then $|x^2y|\le r^3$ and $|y^3|\le r^3$. The triangle inequality consequently gives $|P(x,y)|\le (|a|+|b|)r^3$. The coefficient matters: it is part of the constant in the final estimate.

The same reasoning gives a general useful rule. A polynomial whose terms all have total degree at least $d$ can be bounded by a constant times $r^d$ when $0<r\le1$. Indeed, a term of degree $k$ is bounded by its coefficient magnitude times $r^k$, and $r^k\le r^d$ when $k\ge d$ and $r\le1$. Adding the finitely many term bounds gives the result.

Takeaway: Near the origin, a monomial of total degree $k$ is bounded by a constant times $r^k$. For a finite sum, add the coefficient magnitudes to obtain a valid constant. When comparing different powers of $r$, specify that $0<r\le1$ if you use $r^k\le r^d$ for $k\ge d$.

Worked Example: A Polynomial Quotient

Evaluate the limit $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x^4+3x^2y^2+2y^4}{x^2+y^2}$. The denominator is nonzero away from the origin.

Each degree-four term has a direct bound: $|x^4|=|x|^4\le r^4$, $|x^2y^2|=|x|^2|y|^2\le r^4$, and $|y^4|\le r^4$. Applying the triangle inequality, including the coefficients, yields

|x^4+3x^2y^2+2y^4| \le |x^4|+3|x^2y^2|+2|y^4| \le r^4+3r^4+2r^4 =6r^4.

The denominator equals $r^2$. Dividing the numerator bound by this positive quantity gives

\left|\frac{x^4+3x^2y^2+2y^4}{x^2+y^2}\right| \le \frac{6r^4}{r^2} =6r^2.

Since $6r^2\to0$, the Squeeze Theorem shows that the limit is zero. To check the epsilon-delta condition directly, choose $\delta=\sqrt{\varepsilon/6}$. If $0<r<\delta$, then $|f(x,y)|\le6r^2<6\delta^2=\varepsilon$. The estimate therefore proves the limit.

Worked Example: Using the Sharper Mixed-Product Bound

Evaluate $\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}$, for nonzero $(x,y)$.

Here the denominator is $r$. Use the sharper product inequality $|xy|\le r^2/2$, and divide by $r>0$:

\left|\frac{xy}{\sqrt{x^2+y^2}}\right| =\frac{|xy|}{r} \le\frac{r^2/2}{r} =\frac r2.

The bound $r/2$ tends to zero, so the limit is zero. For any $\varepsilon>0$, take $\delta=2\varepsilon$. Then $0<r<\delta$ implies $|xy|/r\le r/2<\delta/2=\varepsilon$, as required. This also confirms the scale of the estimate: when $|x|=|y|$, the sharper product inequality is an equality.

A Common Inequality Error

A frequent mistake is to reverse the coordinate inequality and write $r\le |x|$. The correct relation is $|x|\le r$: the distance includes contributions from both coordinates and can therefore be larger than either coordinate magnitude. For example, at $(x,y)=(1,2)$,

|x|=1, \qquad r=\sqrt{1^2+2^2}=\sqrt5, \qquad 1<\sqrt5.

Thus $r\le|x|$ is false at this point. The correction is to use $x^2\le x^2+y^2$, which gives $|x|\le r$ after taking square roots. A second related error is to assume that a denominator is at least $r^2$ when only an upper bound has been shown. Before dividing an estimate, verify the direction of the denominator inequality and that the denominator is positive on the points under consideration.

Why These Estimates Matter

A multivariable limit concerns all ways of approaching a point. A coordinate estimate replaces the direction-dependent coordinates with the single nonnegative distance $r$, so one bound can control every nearby point at once. This is why these inequalities are particularly effective alongside the Squeeze Theorem.

The main work is often deciding whether the resulting power of $r$ tends to zero. In a quotient, bound the numerator and denominator separately, then divide only using a valid lower bound for the denominator. A numerator bound alone cannot settle a quotient if the denominator may also become small. Keep track of constants and restrictions such as $r\le1$, and the estimate will state precisely what it proves.

Check Your Understanding

Use the distance $r=\sqrt{x^2+y^2}$ to justify each estimate or conclusion.

Explain why $|x|\le r$ and $|y|\le r$.
What bound does the coordinate-power theorem give for $|x|^3|y|^2$?
Prove that $|xy|\le r^2/2$ using a nonnegative square.
Use a distance bound to evaluate $\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^2}$.
Why must a lower bound for a positive denominator be checked before dividing an inequality by it?