Why Check Components?
A vector-valued function assigns a vector to each input. For example, a function of a real parameter \(t\) might describe a point moving through three-dimensional space. To determine where that vector approaches as \(t\) approaches a value, it is enough to examine each coordinate separately—but every coordinate must have a limit.
This component-by-component method is useful because limits of real-valued functions are familiar from earlier calculus. The key fact is that in a finite-dimensional vector, all coordinates must settle to their respective limits for the vector itself to settle.
For this tutorial, distance is measured with the Euclidean norm: if \(\mathbf{v}=(v_1,\ldots,v_n)\), then \(\|\mathbf{v}\|=\sqrt{v_1^2+\cdots+v_n^2}\). Thus the definition of the vector limit says that for every \(\varepsilon>0\), inputs sufficiently close to \(a\) make \(\|\mathbf{r}(t)-\mathbf{L}\|<\varepsilon\). The value of the function at \(a\), if it is defined, does not determine this limit; the limit depends on nearby inputs.
The Componentwise Limit Theorem
The requirement is that every component limit exists. One coordinate that fails to approach a single finite value is enough to prevent a vector limit.
Why the Theorem Works
The proof follows from two comparisons between the Euclidean distance and the size of an individual coordinate. For any vector \(\mathbf{v}=(v_1,\ldots,v_n)\), each \(|v_i|\leq \|\mathbf{v}\|\). Also, if every coordinate has absolute value less than \(\delta\), then \(\|\mathbf{v}\|<\sqrt{n}\,\delta\). These inequalities establish both directions.
Suppose \(\|\mathbf{r}(t)-\mathbf{L}\|\) approaches zero. For every \(i\), \(|f_i(t)-L_i|\leq \|\mathbf{r}(t)-\mathbf{L}\|\). Therefore, making the vector distance less than any given positive tolerance also makes the difference in each component less than that tolerance. Each \(f_i(t)\) approaches \(L_i\).
Suppose every \(f_i(t)\) approaches \(L_i\). Given \(\varepsilon>0\), set \(\delta=\varepsilon/\sqrt{n}\). For each component, its limit gives a neighborhood of \(a\) in which \(|f_i(t)-L_i|<\delta\). There are only finitely many components, so we can take the smallest of their neighborhood widths. For inputs in that common neighborhood, \(\|\mathbf{r}(t)-\mathbf{L}\|<\sqrt{n}\,\delta=\varepsilon\). Thus the vector limit is \(\mathbf{L}\).
This argument depends on having finitely many components: we can find a common neighborhood for all of them, and the factor \(\sqrt{n}\) is finite.
Finding a Limit One Coordinate at a Time
When the theorem applies, calculate the scalar limits separately and then collect them in the same order as the components. If a component formula has a removable factor, simplify it for nearby inputs; the limit concerns those inputs, not necessarily the original formula's value at \(a\).
Worked Example: A Limit in Three Components
Find \(\lim_{t\to 1}\mathbf{r}(t)\), where, for \(t\neq 1\), \(\mathbf{r}(t)=\left(\frac{t^2-1}{t-1},(t-1)(t+2),\frac{1}{t+1}\right)\).
First find the first component limit. For \(t\neq 1\), factor and cancel:
Therefore, \(\lim_{t\to 1}f_1(t)=1+1=2\). For the second component, substitution gives \(\lim_{t\to 1}(t-1)(t+2)=(1-1)(1+2)=0\). For the third, \(\lim_{t\to 1}\frac{1}{t+1}=\frac{1}{1+1}=\frac12\). All three component limits exist, so the theorem gives
Check: The first coordinate approaches \(2\), the second approaches \(0\), and the third approaches \(\frac12\). These are exactly the three coordinates of the stated vector limit. The original first-component formula is undefined at \(t=1\), but that does not prevent its limit from existing.
When One Component Prevents a Limit
It is not enough for most coordinates to behave well. If even one coordinate has no limit, the componentwise theorem says the vector limit does not exist. Oscillation provides a direct way to demonstrate this: find two sequences of inputs approaching the same point for which that coordinate approaches different values.
Worked Example: An Oscillating Component
For \(t\neq 0\), let \(\mathbf{q}(t)=(t^2,\sin(1/t),3+t)\). Determine whether \(\lim_{t\to 0}\mathbf{q}(t)\) exists.
The first and third component limits are \(\lim_{t\to 0}t^2=0\) and \(\lim_{t\to 0}(3+t)=3\). For the middle component, consider the two sequences
As \(k\to\infty\), both \(t_k\) and \(s_k\) approach \(0\). But
The middle component therefore cannot approach a single value as \(t\to 0\). By the componentwise limit theorem, the vector limit does not exist.
Check: The first and third coordinates approach \(0\) and \(3\) along both sequences, while the second coordinate is \(1\) along the first sequence and \(-1\) along the second. The two different behaviors confirm that the vector cannot approach one fixed vector.
Continuity at a Point
Combining this definition with the componentwise limit theorem gives the continuity test: \(\mathbf{r}\) is continuous at \(a\) if and only if every component \(f_i\) is continuous at \(a\). In particular, each component must be defined at \(a\), have a limit there, and have that limit equal to its value at \(a\). A mismatch in even one component makes the vector function discontinuous.
Worked Example: A Limit Exists but Continuity Fails
Define \(\mathbf{p}(t)=(p_1(t),p_2(t),p_3(t))\) near \(t=1\), where
For \(t\neq 1\), \(p_2(t)=t-1\). Thus the component limits as \(t\to 1\) are
All component limits exist, so \(\lim_{t\to 1}\mathbf{p}(t)=(2,0,5)\). However, evaluating at \(t=1\) gives \(\mathbf{p}(1)=(2,0,6)\). Since \((2,0,5)\neq(2,0,6)\), the vector function is not continuous at \(1\).
Check: The first two limit coordinates match \(p_1(1)=2\) and \(p_2(1)=0\). The third limit is \(5\), while \(p_3(1)=6\). That single mismatch confirms the discontinuity.
A Common Mistake
A tempting but incorrect approach is to check only the components that appear simple, or to conclude that a vector limit exists because several components have limits. For example, in the function \(\mathbf{q}\) above, the first and third coordinates do have limits at \(0\). That does not establish a vector limit: the middle coordinate oscillates. The correction is to check every component, and to state “the vector limit does not exist” as soon as one component fails to have a limit.
Another common error is to confuse the limit with the value at the point. In the example involving \(\mathbf{p}\), all the component limits exist, so the vector limit exists. Yet continuity fails because the limit vector differs from \(\mathbf{p}(1)\). Always test these as separate conditions.
Why the Component Test Matters
Vector-valued functions are often displayed as ordered lists of scalar functions. The component test turns questions about the whole vector into familiar scalar limit and continuity questions, without losing information. It also gives a reliable way to diagnose failure: an oscillating, unbounded, or otherwise nonconvergent component rules out the corresponding vector limit. Once the limits are known, comparing their coordinates with the function’s value determines continuity.
Check Your Understanding
For each question, examine all components and distinguish the limit from the value at the point where appropriate.
- If \(\mathbf{r}(t)=(t^2,3t-1)\), what is \(\lim_{t\to 2}\mathbf{r}(t)\)?
- Why is a vector limit impossible if one component has different limits along two sequences approaching the same input?
- For \(t\neq 0\), let \(\mathbf{u}(t)=(t,\cos(1/t))\). Does \(\lim_{t\to 0}\mathbf{u}(t)\) exist? Explain using the components.
- A vector function has component limits \((1,-2,4)\) at \(a\) and has value \((1,-2,5)\) at \(a\). Is it continuous there?
- State the componentwise condition equivalent to continuity of a vector-valued function at a point.
