Continuity of Vector-Valued Functions

Continuity of a Vector-Valued Function

A vector-valued function traces a path through space as its parameter changes. Continuity means the path has no break at the point being considered: as the parameter approaches that point, the vector approaches the value assigned there. The test can be carried out one coordinate at a time, which is especially useful for piecewise-defined functions.

For a function with values in \(\mathbb{R}^n\), write its components as \(f_1,\ldots,f_n\). The key is to check each component at the same parameter value. If even one component fails the continuity test, the vector-valued function is not continuous there.

Definition: Let \(I\) be an interval, let \(\mathbf{r}:I\to\mathbb{R}^n\), and let \(a\in I\). The function \(\mathbf{r}\) is continuous at \(a\) if the limit of \(\mathbf{r}(t)\) as \(t\) approaches \(a\) through \(I\) exists and equals \(\mathbf{r}(a)\). If \(\mathbf{r}(t)=\langle f_1(t),\ldots,f_n(t)\rangle\), the functions \(f_1,\ldots,f_n\) are its component functions.

At an interior point of \(I\), the limit in this definition is a two-sided limit. If \(a\) is an endpoint of \(I\), it is the one-sided limit from within \(I\). This endpoint convention is essential: a function defined only for \(t\geq a\) is tested as \(t\to a^+\), not by asking for values at parameters outside its domain.

The Component Test

The earlier material on vector limits establishes that a vector limit can be evaluated by taking the limits of its components. Applying that result to the definition of continuity gives the practical test below.

Theorem: Let \(\mathbf{r}(t)=\langle f_1(t),\ldots,f_n(t)\rangle\) be defined on an interval \(I\), and let \(a\in I\). Then \(\mathbf{r}\) is continuous at \(a\) if and only if every component \(f_i\) is continuous at \(a\), using a one-sided limit when \(a\) is an endpoint of \(I\).

To see how the test follows, suppose first that every component is continuous at \(a\). Then the limit of each \(f_i(t)\) is \(f_i(a)\). The component limit rule gives the vector limit as the vector of those component limits, which is exactly \(\langle f_1(a),\ldots,f_n(a)\rangle=\mathbf{r}(a)\). Thus \(\mathbf{r}\) is continuous at \(a\). Conversely, if \(\mathbf{r}\) is continuous at \(a\), its limit is \(\mathbf{r}(a)\). The component limit rule then identifies the limit of each \(f_i\) as \(f_i(a)\), so every component is continuous at \(a\). This proves both directions.

For continuity on an interval, apply the point test at every point. On an open interval there are no included endpoints to check. On a closed interval, check two-sided continuity at interior points and one-sided continuity from within the interval at each endpoint. On a half-open interval, check only the endpoint that belongs to the interval.

Worked Example: Testing Continuity at an Interior Point

Let \(\mathbf{r}(t)=\langle t^2-3t+4,\;2t+\sin(\pi t)\rangle\). Test continuity at \(t=1\).

The first component is \(f_1(t)=t^2-3t+4\). Its limit and value are

\lim_{t\to1}f_1(t)=1^2-3(1)+4=1-3+4=2, \qquad f_1(1)=1^2-3(1)+4=2.

The second component is \(f_2(t)=2t+\sin(\pi t)\). Using continuity of the sine function,

\lim_{t\to1}f_2(t)=2(1)+\sin(\pi\cdot1)=2+\sin(\pi)=2, \qquad f_2(1)=2(1)+\sin(\pi)=2.

Both component limits equal their values, so the component test gives

\lim_{t\to1}\mathbf{r}(t)=\langle 2,2\rangle =\mathbf{r}(1).

Therefore \(\mathbf{r}\) is continuous at \(t=1\). Substituting \(t=1\) into the original function confirms the checked value: \(\mathbf{r}(1)=\langle 2,2\rangle\).

Checking a Piecewise Junction

For a piecewise-defined function, the formulas on either side of a junction may differ. Continuity at the junction is not determined by whether the formulas look alike; it is determined by whether each component approaches the value assigned at the junction. At an interior junction, check the left-hand and right-hand limits of each component and compare them with its value there.

Worked Example: A Piecewise Function Continuous at the Junction

Define \(\mathbf{r}:\mathbb{R}\to\mathbb{R}^3\) by

\mathbf{r}(t)= \begin{cases} \langle t+2,\;3t+1,\;1-t\rangle, & t<0,\\ \langle 2-t^2,\;t^2+1,\;1+t\rangle, & t\geq0. \end{cases}

Test continuity at the junction \(t=0\). The value comes from the \(t\geq0\) branch:

\mathbf{r}(0)=\langle 2-0^2,\;0^2+1,\;1+0\rangle =\langle 2,1,1\rangle.

From the left, use the \(t<0\) formulas and take each limit:

\lim_{t\to0^-}(t+2)=2,\qquad \lim_{t\to0^-}(3t+1)=1,\qquad \lim_{t\to0^-}(1-t)=1.

Thus the left-hand vector limit is \(\langle2,1,1\rangle\). From the right, use the \(t\geq0\) formulas:

\lim_{t\to0^+}(2-t^2)=2,\qquad \lim_{t\to0^+}(t^2+1)=1,\qquad \lim_{t\to0^+}(1+t)=1.

The right-hand vector limit is also \(\langle2,1,1\rangle\). Both sides agree with the assigned value, so the two-sided limit exists and

\lim_{t\to0}\mathbf{r}(t) =\langle2,1,1\rangle =\mathbf{r}(0).

Therefore \(\mathbf{r}\) is continuous at \(0\). Each branch consists of polynomial components, so it is also continuous away from the junction. The junction check is the only additional check needed to establish continuity on all of \(\mathbb{R}\).

One Component Can Break Continuity

The component test is an “every component” test. Several components can be continuous while one is not, but that does not make the vector function continuous. In particular, matching the vector value from one side is not enough: at an interior point, both one-sided limits must agree with the function value.

Worked Example: A Piecewise Function Discontinuous at the Junction

Let

\mathbf{q}(t)= \begin{cases} \langle t^2,\;t+1,\;2t\rangle, & t<1,\\ \langle t^2,\;2t+1,\;2t\rangle, & t\geq1. \end{cases}

The assigned value is taken from the second branch:

\mathbf{q}(1)=\langle 1^2,\;2(1)+1,\;2(1)\rangle =\langle1,3,2\rangle.

From the left, the component limits are

\lim_{t\to1^-}t^2=1,\qquad \lim_{t\to1^-}(t+1)=2,\qquad \lim_{t\to1^-}2t=2.

Therefore \(\lim_{t\to1^-}\mathbf{q}(t)=\langle1,2,2\rangle\), which differs from \(\mathbf{q}(1)=\langle1,3,2\rangle\). The second component alone fails the continuity test: its left-hand limit is \(2\), while its value is \(3\). Hence \(\mathbf{q}\) is not continuous at \(1\). The first and third components do have matching left limits and values, but that does not repair the failure in the second component.

Continuity on an Interval with Endpoints

At an endpoint, continuity is relative to the interval. For example, if the domain is \([0,4]\), the function need not be defined to the left of \(0\) or to the right of \(4\). The appropriate tests use only values from within \([0,4]\).

Worked Example: Continuity on a Closed Interval

Consider \(\mathbf{s}:[0,4]\to\mathbb{R}^2\) defined by \(\mathbf{s}(t)=\langle\sqrt{t},t^2\rangle\). At the left endpoint,

\lim_{t\to0^+}\sqrt{t}=0,\qquad \lim_{t\to0^+}t^2=0,\qquad \mathbf{s}(0)=\langle\sqrt0,0^2\rangle=\langle0,0\rangle.

The right-hand component limits equal the components of \(\mathbf{s}(0)\), so \(\mathbf{s}\) is continuous at \(0\) relative to its domain. At the right endpoint,

\lim_{t\to4^-}\sqrt{t}=2,\qquad \lim_{t\to4^-}t^2=16,\qquad \mathbf{s}(4)=\langle\sqrt4,4^2\rangle=\langle2,16\rangle.

Thus \(\mathbf{s}\) is continuous at \(4\) relative to \([0,4]\). For every interior point \(0<t<4\), both component functions are continuous, so the component test gives continuity there as well. Therefore \(\mathbf{s}\) is continuous on the entire interval \([0,4]\). At both endpoints, substituting the endpoint into the function confirms that the one-sided limit equals the assigned vector value.

A Common Mistake

A frequent error at a piecewise junction is to use the branch that defines the value at the junction to calculate both one-sided limits. In the example \(\mathbf{q}\), the value \(\mathbf{q}(1)=\langle1,3,2\rangle\) comes from the \(t\geq1\) branch. A mistaken check might substitute \(t=1\) into that branch and conclude that the limit is \(\langle1,3,2\rangle\), then declare continuity.

That calculation checks the value, not the left-hand limit. For \(t<1\), the second component is \(t+1\), so its left-hand limit is \(1+1=2\), not \(3\). The correction is to calculate the left-hand limit using the left-hand formula, calculate the right-hand limit using the right-hand formula, and then compare both with the assigned value.

Takeaway: A vector-valued function is continuous at a point exactly when every component is continuous there. At an interior piecewise junction, check both one-sided limits for every component; at an interval endpoint, use the one-sided limit from within the domain.

Check Your Understanding

Use component limits and function values to answer these questions.

If \(\mathbf{r}(t)=\langle f(t),g(t)\rangle\), what must be true of \(f\) and \(g\) for \(\mathbf{r}\) to be continuous at an interior point \(a\)?
For a piecewise vector function with a junction at \(t=0\), which formulas should be used to calculate the left-hand limits?
If one component has a left-hand limit at a junction that differs from its value there, can the vector function be continuous there? Explain.
For a function defined on \([2,5]\), which one-sided limit is used to test continuity at \(t=2\)?
What must be checked to establish continuity of a vector-valued function on a closed interval?