Vector Limits from Component Limits

Why Take a Vector Limit One Coordinate at a Time?

A parametrized curve assigns a position in the plane or in space to each parameter value. As the parameter approaches a particular value, the curve may approach a definite point. To identify that point, track each coordinate of the position separately.

This is useful because the coordinates are scalar functions, and scalar limits are familiar from earlier calculus. A two-dimensional curve has two coordinate limits to find; a three-dimensional curve has three. The resulting numbers form the candidate limiting point.

Key idea: A finite limit of a vector-valued function is determined coordinate by coordinate. If every coordinate approaches a finite number, those numbers are the coordinates of the limiting vector.

Definition of a Vector Limit from Component Limits

Let \(a\) be a parameter value, and suppose a vector-valued function is defined for parameter values in some punctured interval around \(a\). In two dimensions, write the function as \(\mathbf{r}(t)=\langle f(t),g(t)\rangle\). In three dimensions, write it as \(\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle\). The component functions are the scalar functions that provide the coordinates.

Definition: If the finite scalar limits of all the component functions exist as \(t\) approaches \(a\), then the vector limit is the vector whose coordinates are those scalar limits. In two dimensions, \(\lim_{t\to a}\mathbf{r}(t)=\langle L_1,L_2\rangle\) when \(\lim_{t\to a}f(t)=L_1\) and \(\lim_{t\to a}g(t)=L_2\). In three dimensions, the corresponding definition is \(\lim_{t\to a}\mathbf{r}(t)=\langle L_1,L_2,L_3\rangle\) when the three component limits are \(L_1,L_2,L_3\).

The limit concerns parameter values near \(a\), not necessarily the value of the function at \(a\). The function might not be defined at \(a\), or its value there might differ from the limiting vector. What matters is how the coordinates behave as \(t\) approaches \(a\).

Each component limit must be finite for a finite vector limit to exist. If a component has no limit, or grows without bound, there is no finite vector that the curve approaches in the ordinary coordinate-by-coordinate sense.

How Coordinate Limits Relate to Distance

The geometric meaning of a vector limit is that the curve’s position gets arbitrarily close to a fixed vector. In two dimensions, the Euclidean distance between \(\mathbf{r}(t)=\langle f(t),g(t)\rangle\) and a proposed limit \(\mathbf{L}=\langle L_1,L_2\rangle\) is

\|\mathbf{r}(t)-\mathbf{L}\| = \sqrt{(f(t)-L_1)^2+(g(t)-L_2)^2}.

In three dimensions, the distance has an additional squared coordinate difference under the square root. If this distance tends to zero, then each individual coordinate difference must tend to zero: the absolute value of any coordinate difference is no greater than the full distance. For example, in three dimensions,

|f(t)-L_1| \leq \sqrt{(f(t)-L_1)^2+(g(t)-L_2)^2+(h(t)-L_3)^2}.

So a curve cannot approach a vector while one of its coordinates stays a definite distance away from the proposed coordinate. The component definition gives a practical way to find the candidate point. The distance formula explains what that point means geometrically.

A Reliable Coordinate-by-Coordinate Process

Write the components.
Separate the vector-valued function into its scalar coordinate functions.

Find each scalar limit.
Evaluate what every coordinate approaches as the parameter approaches the specified value.

Assemble the candidate vector.
Place the finite coordinate limits in their original order.

Check that every coordinate qualifies.
If any coordinate has no finite limit, do not report a finite vector limit.

This process does not require the coordinates to approach their limits at the same rate. One coordinate may settle down quickly while another approaches its value more slowly. The limiting vector is still determined by their respective limits.

Worked Examples

Worked Example: A Two-Dimensional Polynomial Curve

Find \(\lim_{t\to 2}\mathbf{r}(t)\) for \(\mathbf{r}(t)=\langle t^2-1,3t+2\rangle\).

The first coordinate is \(t^2-1\). As \(t\to2\), it approaches \(2^2-1=4-1=3\). The second coordinate is \(3t+2\), which approaches \(3(2)+2=6+2=8\). Both coordinate limits are finite, so the vector limit is

\lim_{t\to2}\mathbf{r}(t)=\langle 3,8\rangle.

Check the result by measuring the difference from the proposed vector. We have

\mathbf{r}(t)-\langle3,8\rangle = \langle t^2-4,3t-6\rangle = \langle(t-2)(t+2),3(t-2)\rangle.

Thus the squared distance is \((t-2)^2(t+2)^2+9(t-2)^2\), which tends to \(0+0=0\) as \(t\to2\). The curve approaches \(\langle3,8\rangle\), as the component calculation predicts.

Worked Example: A Two-Dimensional Curve with Trigonometric Coordinates

Find \(\lim_{t\to0}\mathbf{r}(t)\) for \(t\ne0\), where \(\mathbf{r}(t)=\left\langle\frac{\sin t}{t},t\cos t\right\rangle\).

For the first coordinate, use the standard scalar limit \(\lim_{t\to0}\frac{\sin t}{t}=1\). For the second coordinate, \(\cos t\to1\) and \(t\to0\), so \(t\cos t\to0\). Therefore, the component limits give

\lim_{t\to0}\mathbf{r}(t)=\langle1,0\rangle.

To check, subtract the candidate: \(\mathbf{r}(t)-\langle1,0\rangle=\left\langle\frac{\sin t}{t}-1,t\cos t\right\rangle\). The first difference tends to \(0\) by the stated scalar limit. The second tends to \(0\) because \(|t\cos t|\leq |t|\) and \(|t|\to0\). Hence the squared distance from the curve to \(\langle1,0\rangle\) tends to \(0^2+0^2=0\).

Worked Example: A Three-Dimensional Curve

Find \(\lim_{t\to0}\mathbf{r}(t)\) for \(\mathbf{r}(t)=\langle\sqrt{4+t},\frac{\sin(2t)}{t},t^2e^t\rangle\), with \(t\) sufficiently close to \(0\) and \(t\ne0\).

For the first coordinate, continuity of the square-root function at \(4\) gives \(\sqrt{4+t}\to\sqrt4=2\). For the second, rewrite the expression as \(2\frac{\sin(2t)}{2t}\). Since \(2t\to0\), the standard limit \(\frac{\sin u}{u}\to1\) gives \(2\frac{\sin(2t)}{2t}\to2(1)=2\). For the third coordinate, \(t^2\to0\) and \(e^t\to1\), so \(t^2e^t\to0(1)=0\). Therefore,

\lim_{t\to0}\mathbf{r}(t)=\langle2,2,0\rangle.

For a distance check, the coordinate differences from this vector are \(\sqrt{4+t}-2\), \(\frac{\sin(2t)}{t}-2\), and \(t^2e^t\). Each tends to \(0\) by the scalar limits just established. The squared Euclidean distance is the sum of their squares, so it tends to \(0^2+0^2+0^2=0\). This verifies the proposed limiting point.

When a Vector Limit Does Not Exist

The component method is also useful for detecting failure. If even one coordinate fails to approach a finite number, the entire vector cannot approach a finite limiting point. Oscillation is one way a scalar component can fail to have a limit.

Worked Example: An Oscillating Coordinate Prevents a Limit

Consider \(\mathbf{r}(t)=\langle t,\sin(1/t)\rangle\) as \(t\to0\), with \(t\ne0\). The first coordinate approaches \(0\). The second coordinate, however, does not approach a single number.

To see this, take \(t_n=\frac{1}{\pi/2+2\pi n}\) for positive integers \(n\). Then \(t_n\to0\), and \(\sin(1/t_n)=\sin(\pi/2+2\pi n)=1\). Now take \(s_n=\frac{1}{3\pi/2+2\pi n}\). Then \(s_n\to0\), but \(\sin(1/s_n)=\sin(3\pi/2+2\pi n)=-1\). Since the second coordinate approaches different values along these two sequences, it has no limit as \(t\to0\). Therefore, \(\mathbf{r}(t)\) has no finite vector limit.

Common Mistake: Guessing That Oscillation Disappears

A tempting but incorrect calculation for the preceding curve is

\lim_{t\to0}\langle t,\sin(1/t)\rangle \overset{\text{wrong}}{=} \langle0,0\rangle,

on the grounds that the first coordinate tends to zero and the sine function takes values near zero. But \(\sin(1/t)\) does not stay near zero as \(t\) approaches zero; it continues to take the values \(1\) and \(-1\) along parameter sequences approaching zero. The corrected conclusion is that the second component limit does not exist, so there is no finite vector limit.

Pitfall to avoid: Finding a limit for some coordinates does not compensate for a coordinate that has no finite limit. Check every component before assembling a vector.

Why This View Matters for Space Curves

A space curve is recorded as three coordinate functions. Its limiting point describes where the curve’s position goes as the parameter approaches a specified value. That point can be found without trying to sketch the curve: evaluate the three scalar limits and put the results in order.

This coordinate-by-coordinate approach also keeps the geometry precise. A curve may pass through a point at one parameter value, approach a point it never reaches, or fail to settle toward any finite point. In each case, the component limits describe the behavior of its coordinates, and the vector limit exists only when those coordinate behaviors fit together as finite limits.

Check Your Understanding

For each question, work with the component functions and state whether a finite vector limit exists.

If \(\mathbf{r}(t)=\langle t^2,2t-1\rangle\), what is \(\lim_{t\to3}\mathbf{r}(t)\)?
If \(\mathbf{r}(t)=\langle\cos t,e^t,t\rangle\), what is \(\lim_{t\to0}\mathbf{r}(t)\)?
Why is it not enough to find finite limits for only two of the three coordinates of a space curve?
For \(\mathbf{r}(t)=\langle t,\sin(1/t)\rangle\), which component prevents a finite limit as \(t\to0\)?
In the Euclidean distance formula, why must each coordinate difference tend to zero if the distance to a proposed limiting vector tends to zero?