The Squeeze Theorem for Multivariable Limits

Why Bounds Can Determine a Limit

Some multivariable functions are awkward to simplify near a point. Their formulas may contain several variables in a way that makes direct substitution or algebraic rearrangement unhelpful. It can be enough, however, to show that their values are trapped between two simpler functions that approach the same number.

The idea is the same as for limits of functions of one variable, but the meaning of “near the point” is now determined by distance in the plane or in higher-dimensional space. The bounds must hold at every point sufficiently close to the target, except possibly at the target itself.

Key idea: If a function is bounded below and above by functions that both approach \(L\), then the function itself approaches \(L\).

Statement of the Squeeze Theorem

For a point \(\mathbf{a}\) in \(\mathbb{R}^n\), write \(\|\mathbf{x}-\mathbf{a}\|\) for the Euclidean distance from \(\mathbf{x}\) to \(\mathbf{a}\). A punctured neighborhood of \(\mathbf{a}\) consists of points satisfying \(0<\|\mathbf{x}-\mathbf{a}\|<\delta\) for some \(\delta>0\). The point itself is excluded because the limit concerns nearby values, not necessarily the value at \(\mathbf{a}\).

Theorem (Squeeze Theorem for multivariable limits): Suppose \(f\), \(g\), and \(h\) are defined on a punctured neighborhood of \(\mathbf{a}\), and for every \(\mathbf{x}\) in that neighborhood, \(g(\mathbf{x})\leq f(\mathbf{x})\leq h(\mathbf{x})\). If \(\lim_{\mathbf{x}\to\mathbf{a}}g(\mathbf{x})=L\) and \(\lim_{\mathbf{x}\to\mathbf{a}}h(\mathbf{x})=L\), then \(\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})=L\).

The bounds need to apply throughout some punctured neighborhood, not just along selected paths. The two bounding limits must also be the same finite number. This theorem establishes the limit of \(f\); it does not require \(f\) to be defined at \(\mathbf{a}\).

Why the Theorem Works

The definition of a multivariable limit says that, for every tolerance \(\varepsilon>0\), all function values sufficiently close to the target point must lie within \(\varepsilon\) of \(L\). Since \(g\) and \(h\) both approach \(L\), they eventually lie inside that same interval. The bounds then force \(f\) inside it as well.

Choose a tolerance.
Let \(\varepsilon>0\). Since \(g(\mathbf{x})\to L\), there is \(\delta_1>0\) such that \(0<\|\mathbf{x}-\mathbf{a}\|<\delta_1\) implies \(|g(\mathbf{x})-L|<\varepsilon\). Thus \(L-\varepsilon<g(\mathbf{x})<L+\varepsilon\).

Apply the other limit.
Since \(h(\mathbf{x})\to L\), there is \(\delta_2>0\) such that \(0<\|\mathbf{x}-\mathbf{a}\|<\delta_2\) implies \(L-\varepsilon<h(\mathbf{x})<L+\varepsilon\).

Combine the bounds.
Take \(\delta=\min(\delta_1,\delta_2,\delta_0)\), where \(\delta_0\) is a radius on which \(g\leq f\leq h\) holds. Then \(0<\|\mathbf{x}-\mathbf{a}\|<\delta\) gives \(L-\varepsilon<g(\mathbf{x})\leq f(\mathbf{x})\leq h(\mathbf{x})<L+\varepsilon\). Consequently \(|f(\mathbf{x})-L|<\varepsilon\).

This proves the limit directly from its definition: for every \(\varepsilon>0\), the chosen \(\delta>0\) ensures that points within distance \(\delta\), other than \(\mathbf{a}\), have function values within \(\varepsilon\) of \(L\).

Using Absolute-Value Bounds

A common application is a bound on the distance between \(f(\mathbf{x})\) and its proposed limit \(L\). If a nonnegative function \(\phi(\mathbf{x})\) satisfies \(\phi(\mathbf{x})\to0\) and \(|f(\mathbf{x})-L|\leq\phi(\mathbf{x})\), then the absolute-value inequality gives both \(L-\phi(\mathbf{x})\leq f(\mathbf{x})\) and \(f(\mathbf{x})\leq L+\phi(\mathbf{x})\). The two bounds approach \(L\), so the Squeeze Theorem applies.

Useful form: If \(|f(\mathbf{x})-L|\leq\phi(\mathbf{x})\) near \(\mathbf{a}\), with \(\phi(\mathbf{x})\geq0\) and \(\phi(\mathbf{x})\to0\) as \(\mathbf{x}\to\mathbf{a}\), then \(f(\mathbf{x})\to L\).

Worked Examples

Worked Example: A Rational Expression Approaching Zero

Find \(\lim_{(x,y)\to(0,0)}\dfrac{x^2y}{x^2+y^2}\). The expression is defined at every point except the origin, which is precisely the region needed for this limit.

Set \(r=\sqrt{x^2+y^2}\), the distance from \((x,y)\) to the origin. At any nonzero point, \(x^2+y^2>0\), and \(0\leq x^2/(x^2+y^2)\leq1\). Therefore

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

As \((x,y)\to(0,0)\), \(r\to0\). Thus the absolute value of the expression is trapped between \(0\) and \(r\), both of which tend to zero. The limit is \(0\).

For an explicit definition-based check, let \(\varepsilon>0\) and choose \(\delta=\varepsilon\). If \(0<r<\delta\), the estimate gives \(\left|\dfrac{x^2y}{x^2+y^2}\right|\leq r<\varepsilon\). This verifies the claimed limit. As a numerical check at \((x,y)=(3/5,4/5)\), the denominator is \(9/25+16/25=1\), and the function value is \(36/125\). The bound gives \(36/125\leq4/5\leq1=r\), as required.

Worked Example: A Two-Sided Estimate with a Nontrivial Sign

Find the limit at the origin of \(f(x,y)=\dfrac{(x^2+y^2)(x^2-y^2)}{1+x^2+y^2}\).

First, \(x^2+y^2\geq0\) and \(1+x^2+y^2>0\), so the denominator never vanishes. Also, \(|x^2-y^2|\leq x^2+y^2\), since \(x^2\) and \(y^2\) are nonnegative. Let \(r=\sqrt{x^2+y^2}\). Then

|f(x,y)| = \frac{r^2|x^2-y^2|}{1+r^2} \leq \frac{r^4}{1+r^2} \leq r^2.

The last inequality follows because \(1+r^2\geq r^2\), and the denominator is positive. As \(r\to0\), the upper bound \(r^2\) tends to zero. Equivalently, \(-r^2\leq f(x,y)\leq r^2\), and both outer expressions tend to zero. Hence the limit is \(0\).

To verify the limit with a choice of \(\delta\), given \(\varepsilon>0\), take \(\delta=\sqrt{\varepsilon}\). If \(0<r<\delta\), then \(|f(x,y)|\leq r^2<\delta^2=\varepsilon\). For a numerical check at \((3/5,4/5)\), \(r^2=1\), \(x^2-y^2=9/25-16/25=-7/25\), and \(1+r^2=2\). Thus \(f=-7/50\), which satisfies \(|f|=7/50\leq1=r^2\).

Worked Example: A Nonnegative Function Bounded by Distance

Consider \(f(x,y)=\dfrac{(x^2+y^2)^{3/2}}{x^2+y^2+|xy|}\) for \((x,y)\ne(0,0)\). We will find its limit at the origin.

Write \(r=\sqrt{x^2+y^2}\). The numerator is \(r^3\), and the denominator is \(r^2+|xy|\geq r^2>0\). Therefore

0\leq f(x,y) =\frac{r^3}{r^2+|xy|} \leq\frac{r^3}{r^2} =r.

Both bounds, \(0\) and \(r\), tend to zero as \(r\to0\). The Squeeze Theorem gives \(\lim_{(x,y)\to(0,0)}f(x,y)=0\).

For the definition check, let \(\varepsilon>0\) and choose \(\delta=\varepsilon\). Whenever \(0<r<\delta\), \(0\leq f(x,y)\leq r<\varepsilon\), so \(|f(x,y)-0|<\varepsilon\). At the test point \((3/5,4/5)\), \(r=1\) and \(|xy|=12/25\). Thus the denominator is \(1+12/25=37/25\), the numerator is \(1\), and \(f=25/37\). This lies between \(0\) and \(1=r\), confirming the estimate at that point.

Limits at Points Other Than the Origin

The same reasoning works at any point \(\mathbf{a}\). One uses the distance \(\|\mathbf{x}-\mathbf{a}\|\), rather than the distance from the origin. The squeeze argument itself does not depend on where the point is located.

Worked Example: Shifting the Center of the Estimate

Find the limit as \((x,y)\to(1,-2)\) of \(f(x,y)=\dfrac{(x-1)^2(y+2)}{(x-1)^2+(y+2)^2}\).

For points other than \((1,-2)\), the denominator is positive. Set \(r=\sqrt{(x-1)^2+(y+2)^2}\). Since \((x-1)^2\leq r^2\),

|f(x,y)| = |y+2|\frac{(x-1)^2}{(x-1)^2+(y+2)^2} \leq |y+2| \leq r.

As \((x,y)\to(1,-2)\), \(r\to0\), so the limit is \(0\). Explicitly, for any \(\varepsilon>0\), choose \(\delta=\varepsilon\); if \(0<r<\delta\), then \(|f(x,y)|\leq r<\varepsilon\). At \((x,y)=(2,-1)\), \(r=\sqrt{2}\), the numerator is \(1\), the denominator is \(2\), and \(f=1/2\). The estimate checks: \(1/2\leq1\leq\sqrt{2}\).

A Common Pitfall: One-Sided Bounds Are Not Enough

It is not enough to show only that \(f(\mathbf{x})\leq\phi(\mathbf{x})\) with \(\phi(\mathbf{x})\to0\). That provides no lower control on \(f\). For example, let \(f(x,y)=-1\) at every point near the origin. Then \(f(x,y)\leq r\) for every \(r\geq0\), and \(r\to0\), but \(f(x,y)\) does not approach zero.

The mistake would be to conclude “\(f\leq r\) and \(r\to0\), so \(f\to0\).” The correction is to establish a two-sided estimate. In this example, \(-1\leq f(x,y)\leq r\), and the lower bound does not approach zero; the squeeze theorem therefore gives no conclusion. A bound such as \(|f|\leq r\), by contrast, supplies both \(-r\leq f\) and \(f\leq r\).

When the Squeeze Theorem Is Useful

The theorem is useful when a complicated formula has a simple size estimate near the point of interest. Often the exact values vary with direction, but a uniform bound in terms of distance still forces all values to approach the same limit. The estimate must hold for every sufficiently nearby point; checking only a line or a handful of paths is not enough to establish it.

Before using the theorem, verify three things:

The proposed bounds are valid on a punctured neighborhood of the point.
The lower and upper bounds both approach the same number.
The inequalities have the correct direction and control both sides of the function’s values.

When the target is \(L\), an estimate on \(|f-L|\) is especially convenient because it immediately gives the required two-sided bounds. The examples show how an estimate by a power of the distance can establish a limit even when the original expression is not well suited to direct substitution.

Check Your Understanding

Use the theorem and the estimates in this tutorial to answer the following questions.

State the two limit conditions and the inequality required to apply the multivariable Squeeze Theorem.
If \(|f(\mathbf{x})-L|\leq\|\mathbf{x}-\mathbf{a}\|^2\) near \(\mathbf{a}\), what is \(\lim_{\mathbf{x}\to\mathbf{a}}f(\mathbf{x})\)?
Why must the inequality \(g\leq f\leq h\) hold throughout a punctured neighborhood rather than just along one path?
For \(f(x,y)=\dfrac{x^2y}{x^2+y^2}\), what distance-based upper bound on \(|f(x,y)|\) proves the limit at the origin?
Give an example showing why \(f(\mathbf{x})\leq\phi(\mathbf{x})\) and \(\phi(\mathbf{x})\to0\) alone do not imply that \(f\to0\).