Approaching a Point with Vector Values
Connectedness concerns how a subset of \(\mathbb{R}^n\) is arranged as a space. Limits address a different question: what value does a function approach as its input approaches a specified point? For a vector-valued function, both the input and output may have several coordinates. The Euclidean norm lets us express closeness in either space with one inequality, while the coordinates provide a useful way to check the resulting limit.
Let \(D\subseteq\mathbb{R}^m\), let \(a\in\mathbb{R}^m\), and let \(f:D\to\mathbb{R}^n\). We say that \(a\) is an accumulation point of \(D\) if every open ball centered at \(a\) contains a point of \(D\) other than \(a\). The value \(a\) need not itself belong to \(D\). Requiring \(a\) to be an accumulation point ensures that inputs in the domain can approach \(a\) without being equal to it.
The condition \(0<\|x-a\|_2\) excludes the input \(x=a\), if it belongs to \(D\). Consequently, the value \(f(a)\) has no bearing on whether this limit exists. The limit describes the values at nearby domain points, not necessarily the value assigned at \(a\). The same definition applies when \(a\notin D\).
The input and output norms are taken in their respective spaces: \(\|x-a\|_2\) is a norm in \(\mathbb{R}^m\), and \(\|f(x)-L\|_2\) is a norm in \(\mathbb{R}^n\). The definition says that inputs sufficiently close to \(a\), but not equal to it, have outputs as close as desired to \(L\).
Uniqueness and a Direct Estimate
A limit, when it exists, cannot depend on which target vector is proposed. This follows from the triangle inequality.
Proof. Suppose that both \(L\) and \(M\) satisfy the definition of the limit. Assume, for contradiction, that \(L\neq M\), and set \(\varepsilon=\|L-M\|_2/3>0\). The limit definitions give positive numbers \(\delta_L\) and \(\delta_M\) such that \[ 0<\|x-a\|_2<\delta_L \quad\Longrightarrow\quad \|f(x)-L\|_2<\varepsilon \] and \[ 0<\|x-a\|_2<\delta_M \quad\Longrightarrow\quad \|f(x)-M\|_2<\varepsilon. \] Because \(a\) is an accumulation point of \(D\), there is an \(x\in D\) with \[ 0<\|x-a\|_2<\min(\delta_L,\delta_M). \] For this \(x\), the triangle inequality implies \[ \|L-M\|_2 \leq \|L-f(x)\|_2+\|f(x)-M\|_2 <2\varepsilon =\frac{2}{3}\|L-M\|_2. \] A positive number cannot be strictly less than two-thirds of itself. This contradiction proves \(L=M\). \(\square\)
Worked Example: Estimating a Polynomial Vector Function
Define \(F:\mathbb{R}^2\to\mathbb{R}^2\) by \[ F(x,y)=(x^2+y^2,\ 3x-2y). \] We show directly that \(\lim_{(x,y)\to(0,0)}F(x,y)=(0,0)\). Write \(r=\sqrt{x^2+y^2}\). By the Cauchy–Schwarz inequality, \[ |3x-2y| \leq \sqrt{13}\sqrt{x^2+y^2} =\sqrt{13}\,r. \] The Euclidean norm of a vector is at most the sum of the absolute values of its coordinates, so \[ \|F(x,y)\|_2 \leq |x^2+y^2|+|3x-2y| \leq r^2+\sqrt{13}\,r. \] Given \(\varepsilon>0\), choose \[ \delta=\min\left(1,\frac{\varepsilon}{1+\sqrt{13}}\right). \] If \(0<r<\delta\), then \(r<1\), so \(r^2\leq r\). Therefore, \[ \|F(x,y)\|_2 \leq r^2+\sqrt{13}\,r \leq (1+\sqrt{13})r <(1+\sqrt{13})\delta \leq\varepsilon. \] This verifies the limit by the definition.
The argument is an example of a standard strategy: estimate the output’s distance from the proposed limit in terms of the input’s distance from the approach point, then choose \(\delta\) to make that bound smaller than \(\varepsilon\). The estimate must apply to every domain point close enough to \(a\), not merely to points on a selected line or curve.
A Sequential Criterion for Limits
Sequences give another way to test a limit. A sequence of inputs approaching \(a\), while avoiding \(a\), should produce a sequence of outputs approaching \(L\). In Euclidean spaces, this condition is not only necessary but sufficient.
Proof. Suppose first that \(\lim_{x\to a,\ x\in D}f(x)=L\), and let \((x_k)\) be a sequence in \(D\setminus\{a\}\) with \(x_k\to a\). Given \(\varepsilon>0\), choose \(\delta>0\) from the limit definition. Since \(x_k\to a\), there is an integer \(K\) such that \(k\geq K\) implies \(\|x_k-a\|_2<\delta\). Also \(x_k\neq a\) for every \(k\). The limit definition therefore gives \[ \|f(x_k)-L\|_2<\varepsilon\qquad(k\geq K). \] Thus \(f(x_k)\to L\).
Conversely, suppose the sequential condition holds, but the limit definition fails. Then there is an \(\varepsilon_0>0\) such that for every \(\delta>0\), some \(x\in D\) satisfies \[ 0<\|x-a\|_2<\delta \quad\text{and}\quad \|f(x)-L\|_2\geq\varepsilon_0. \] For each positive integer \(k\), apply this statement with \(\delta=1/k\) to choose \(x_k\in D\) such that \[ 0<\|x_k-a\|_2<\frac{1}{k} \quad\text{and}\quad \|f(x_k)-L\|_2\geq\varepsilon_0. \] The first inequality shows that \(x_k\to a\), but the second shows that \(f(x_k)\) does not converge to \(L\). This contradicts the assumed sequential condition. Hence the limit definition holds. \(\square\)
Worked Example: Different Paths Rule Out a Limit
For \((x,y)\neq(0,0)\), define \[ G(x,y)=\left(\frac{xy}{x^2+y^2},\frac{x^2}{x^2+y^2}\right). \] Consider the two sequences \[ u_k=\left(\frac{1}{k},0\right), \qquad v_k=\left(\frac{1}{k},\frac{1}{k}\right). \] Both belong to the domain and converge to \((0,0)\). Direct substitution gives, for every \(k\geq1\), \[ G(u_k)=\left(\frac{(1/k)\,0}{(1/k)^2+0^2}, \frac{(1/k)^2}{(1/k)^2+0^2}\right)=(0,1), \] whereas \[ G(v_k)=\left(\frac{1/k^2}{2/k^2},\frac{1/k^2}{2/k^2}\right) =\left(\frac12,\frac12\right). \] The output sequences have different limits. By the Sequential Criterion, \(G\) has no limit at \((0,0)\). Equivalently, if a limit existed, each of these sequences of outputs would have to converge to the same vector.
Checking a few paths can disprove a limit, as in this example, but agreement along a few paths does not prove that a limit exists. The sequential criterion quantifies over every sequence in the domain approaching \(a\). A direct norm estimate is often more efficient for proving a limit because it controls all nearby inputs at once.
Limits and Coordinate Functions
Write \(f(x)=(f_1(x),\ldots,f_n(x))\) and \(L=(L_1,\ldots,L_n)\). The Euclidean norm relates the vector limit to its coordinate limits. In particular, for each index \(i\), \[ |f_i(x)-L_i|\leq\|f(x)-L\|_2. \] In the other direction, \[ \|f(x)-L\|_2 =\left(\sum_{i=1}^{n}|f_i(x)-L_i|^2\right)^{1/2} \leq \sqrt{n}\max_{1\leq i\leq n}|f_i(x)-L_i|. \] These inequalities give a precise equivalence.
Proof. Suppose first that \(f(x)\to L\). Given \(\varepsilon>0\), the vector limit definition provides \(\delta>0\) such that \(0<\|x-a\|_2<\delta\) implies \(\|f(x)-L\|_2<\varepsilon\). For each coordinate \(i\), \[ |f_i(x)-L_i|\leq\|f(x)-L\|_2<\varepsilon. \] Thus \(f_i(x)\to L_i\).
Conversely, suppose every coordinate function tends to its corresponding coordinate of \(L\). Given \(\varepsilon>0\), for each \(i\) choose \(\delta_i>0\) such that \[ 0<\|x-a\|_2<\delta_i \quad\Longrightarrow\quad |f_i(x)-L_i|<\frac{\varepsilon}{\sqrt{n}}. \] There are finitely many coordinates, so \(\delta=\min(\delta_1,\ldots,\delta_n)>0\). If \(x\in D\) and \(0<\|x-a\|_2<\delta\), then every coordinate satisfies the displayed bound. Consequently, \[ \|f(x)-L\|_2 =\left(\sum_{i=1}^{n}|f_i(x)-L_i|^2\right)^{1/2} <\left(\sum_{i=1}^{n}\frac{\varepsilon^2}{n}\right)^{1/2} =\varepsilon. \] This proves that \(f(x)\to L\). \(\square\)
Worked Example: Bounding a Vector Function by the Input Distance
For \((x,y)\neq(0,0)\), define \[ H(x,y)=\left(\frac{x^2y}{x^2+y^2},\frac{xy^2}{x^2+y^2}\right). \] Set \(r=\sqrt{x^2+y^2}\). Since \(x^2\leq r^2\) and \(y^2\leq r^2\), \[ \left|\frac{x^2y}{x^2+y^2}\right|\leq |y|, \qquad \left|\frac{xy^2}{x^2+y^2}\right|\leq |x|. \] Squaring and adding these inequalities yields \[ \|H(x,y)\|_2^2 \leq y^2+x^2 =r^2, \] so \(\|H(x,y)\|_2\leq r\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon\). Whenever \(0<\sqrt{x^2+y^2}<\delta\), \[ \|H(x,y)-(0,0)\|_2=\|H(x,y)\|_2 \leq\sqrt{x^2+y^2} <\varepsilon. \] Therefore \(\lim_{(x,y)\to(0,0)}H(x,y)=(0,0)\).
The coordinate criterion gives another route: the first coordinate is bounded in absolute value by \(|y|\), and the second by \(|x|\), so each coordinate tends to zero. The norm estimate additionally gives one bound for the entire vector.
What the Definition Requires
The definition and the criteria above help avoid several common errors:
- The point need only be approached through the domain. The definition concerns \(x\in D\), not every point in \(\mathbb{R}^m\). If some nearby inputs do not belong to \(D\), they play no role.
- The value at the approach point is excluded. If \(a\in D\), changing \(f(a)\) does not change the limit, because the condition requires \(0<\|x-a\|_2\).
- Coordinate limits must all exist and be compatible. A vector limit is one vector \(L\), so its coordinates are the limits of all the coordinate functions. A single coordinate that fails to have a limit prevents the vector limit from existing.
- Path checks have different roles in the two directions. Two sequences giving incompatible output limits disprove a limit. Matching behavior along selected sequences alone does not establish one.
For finite-dimensional Euclidean spaces, the coordinate criterion allows scalar limit techniques to be applied one component at a time, while the norm definition keeps the meaning of the vector limit explicit. The Sequential Criterion gives a second characterization, and direct estimates remain a reliable way to verify that every sufficiently nearby input has an output close to the proposed limit.
Check Your Understanding
Use the definitions and results in this tutorial to answer each question.
- Why must the point \(a\) be an accumulation point of the domain in the definition of a limit?
- Why does changing \(f(a)\), when \(a\in D\), leave the limit at \(a\) unchanged?
- How does the Sequential Criterion prove that a limit does not exist when two input sequences approaching \(a\) give different output limits?
- Why is it enough to choose one \(\delta\) for all coordinate functions in the proof of the Coordinate Criterion?
- For a function with values in \(\mathbb{R}^n\), how can bounds on its coordinate errors give a bound on its Euclidean error?