Why Check Limits One Component at a Time?
A vector-valued function can describe several changing quantities at once. For instance, its components might give the coordinates of a moving point. To understand the point’s limit, it is natural to examine each coordinate. But why is that enough to establish a limit of the whole vector?
The answer depends on how distance between vectors is measured. In ordinary finite-dimensional Euclidean space, the length of a vector difference is determined by its coordinate differences. We will use that relationship to prove both directions: a vector limit implies limits for all components, and limits for all components imply a vector limit.
Definition of a Vector Limit
Let \(D\) be a subset of \(\mathbb{R}\), let \(a\) be a point that can be approached by points of \(D\) other than \(a\), and let \(\mathbf{f}:D\to\mathbb{R}^n\). Write the function in components as \(\mathbf{f}(t)=(f_1(t),\ldots,f_n(t))\). For \(\mathbf{L}=(L_1,\ldots,L_n)\), the statement \(\lim_{t\to a}\mathbf{f}(t)=\mathbf{L}\) means that the vectors \(\mathbf{f}(t)\) get arbitrarily close to \(\mathbf{L}\) as \(t\) approaches \(a\), without requiring \(t=a\).
Closeness is measured by the Euclidean norm. For \(\mathbf{x}=(x_1,\ldots,x_n)\), this norm is \(\|\mathbf{x}\|=\sqrt{x_1^2+\cdots+x_n^2}\).
This is the usual epsilon-delta definition, with the distance between function values measured by the norm. A component limit \(\lim_{t\to a}f_i(t)=L_i\) uses the same definition with the absolute value \(|f_i(t)-L_i|\) in place of the vector norm.
The Componentwise Limit Theorem
The theorem concerns a finite number of components. Its proof shows why the Euclidean norm makes checking coordinates sufficient, and why a small vector difference forces each coordinate difference to be small.
Proof: A Vector Limit Gives Component Limits
Suppose \(\lim_{t\to a}\mathbf{f}(t)=\mathbf{L}\). Fix a component index \(i\) and a number \(\varepsilon>0\). By the definition of the vector limit, there is a \(\delta>0\) such that \(0<|t-a|<\delta\) implies \(\|\mathbf{f}(t)-\mathbf{L}\|<\varepsilon\).
The square of this norm is the sum of the squares of the coordinate differences. Each squared term is nonnegative, so no individual term can exceed the sum. Consequently, the absolute difference in component \(i\) cannot exceed the norm:
Thus, for the same \(\delta\), whenever \(0<|t-a|<\delta\),
This is exactly the definition of \(\lim_{t\to a}f_i(t)=L_i\). Since the argument holds for every component \(i\), every component limit exists and equals the corresponding coordinate of \(\mathbf{L}\).
Proof: Component Limits Give a Vector Limit
Now suppose that \(\lim_{t\to a}f_i(t)=L_i\) for every \(i=1,\ldots,n\). Let \(\varepsilon>0\). For each component, its limit definition provides a positive number \(\delta_i\) such that \(0<|t-a|<\delta_i\) implies \(|f_i(t)-L_i|<\varepsilon/\sqrt{n}\). Because there are finitely many components, the minimum of these positive numbers is positive. Set \(\delta=\min(\delta_1,\ldots,\delta_n)\).
Whenever \(t\in D\) and \(0<|t-a|<\delta\), every component satisfies its required bound. We can therefore add the squares of those bounds:
Both the norm and \(\varepsilon\) are nonnegative, so \(\|\mathbf{f}(t)-\mathbf{L}\|^2<\varepsilon^2\) implies \(\|\mathbf{f}(t)-\mathbf{L}\|<\varepsilon\). This is the definition of the vector limit. The argument also shows that its value is the vector made from the component limits.
Worked Example: Finding a Limit from Three Components
Worked Example: Finding a Limit from Three Components
Let \(\mathbf{f}(t)=(t+1,\ t^2+t+1,\ 2t-1)\). Find \(\lim_{t\to 1}\mathbf{f}(t)\).
First find each component limit. The first component approaches \(1+1=2\). The second approaches \(1^2+1+1=3\). The third approaches \(2(1)-1=1\). Thus the component limits are \(2\), \(3\), and \(1\). By the componentwise limit theorem,
To check the result directly, subtract the proposed limit from the function value:
Therefore,
As \(t\to1\), the square-root factor approaches \(\sqrt{14}\), while \(|t-1|\) approaches zero. The distance therefore approaches zero, confirming the stated vector limit.
Worked Example: An Explicit Epsilon-Delta Bound
Worked Example: An Explicit Epsilon-Delta Bound
Let \(\mathbf{g}(t)=(3t-2,\ 1-2t)\). Prove that \(\lim_{t\to 2}\mathbf{g}(t)=(4,-3)\).
The components at \(t=2\) give \(3(2)-2=4\) and \(1-2(2)=-3\). For \(t\) near \(2\), the difference from the proposed limit is
Its Euclidean norm is exactly
Given any \(\varepsilon>0\), choose \(\delta=\varepsilon/\sqrt{13}\). If \(0<|t-2|<\delta\), then
This verifies the vector limit from its definition. It also agrees with the theorem: the component differences \(3(t-2)\) and \(-2(t-2)\) each tend to zero.
Worked Example: One Component Can Prevent a Limit
Worked Example: One Component Can Prevent a Limit
Consider \(\mathbf{h}(t)=(\sin(1/t),t)\) for \(t\ne0\). Does \(\mathbf{h}(t)\) have a limit as \(t\to0\)?
The second component has limit \(0\), but the first component does not. To verify this, take \(t_n=1/(\pi/2+2\pi n)\) and \(s_n=1/(3\pi/2+2\pi n)\), for positive integers \(n\). Both are positive and approach zero as \(n\) increases. However,
If the first component had a limit \(L\), its values along both sets of parameter values would have to approach \(L\). The first set has constant value \(1\), so it approaches \(1\); the second has constant value \(-1\), so it approaches \(-1\). A limit cannot equal both \(1\) and \(-1\). Thus the first component has no limit. By the theorem, the vector function has no limit as \(t\to0\), despite its second component having a limit.
A Common Mistake with Coordinate Bounds
When proving that component limits imply a vector limit, it is not enough to bound each coordinate difference by \(\varepsilon\) and then claim that the vector distance is less than \(\varepsilon\). In two dimensions, for example, coordinate differences of \(0.8\varepsilon\) each give the distance
The error is in overlooking how the coordinate differences combine under the norm. The correction is to make each of the \(n\) coordinate differences less than \(\varepsilon/\sqrt{n}\). Their squared contributions then sum to less than \(n(\varepsilon/\sqrt{n})^2=\varepsilon^2\), as in the proof.
Why the Result Matters
The theorem lets us reduce a limit question in \(\mathbb{R}^n\) to finitely many ordinary scalar limit questions, without losing information about existence. It applies because the vector has finitely many components and distance is measured by the Euclidean norm. In particular, every coordinate must approach a definite value; oscillation or failure of a limit in even one coordinate rules out a vector limit.
Once the component limits are known, the value of the vector limit is not an additional choice: its \(i\)th coordinate must be the limit of \(f_i(t)\). This is the precise reason that componentwise computation is valid for finite-dimensional vector-valued functions.
Check Your Understanding
Use the definition and the componentwise limit theorem to answer these questions.
- Why does \(|f_i(t)-L_i|\leq\|\mathbf{f}(t)-\mathbf{L}\|\) hold for every component?
- In the proof from component limits, why is it possible to choose one \(\delta\) that works for all components?
- For a function into \(\mathbb{R}^4\), what bound on each coordinate difference is sufficient to guarantee a vector distance less than \(\varepsilon\)?
- If one component has no limit at \(a\), what can you conclude about the vector limit there?
- Why does the proof from component limits rely on having finitely many components?