Limit Laws for Vector-Valued Functions

Why Limit Laws Matter

Vector-valued functions describe quantities with several components, such as the position of a moving point in space. When two such quantities approach limits, we often need the limit of their sum or difference. A scalar factor may also change the size or direction of a vector quantity as the parameter varies.

Limit laws make these calculations systematic: under stated conditions, the limit of an operation on functions can be found by performing that operation on their limits. The conditions matter. In particular, the scalar-factor law requires finite limits for both the scalar function and the vector function.

Setup and Notation

Let $t$ be a real parameter and let $a$ be a point approached by $t$. A vector-valued function $\mathbf{u}$ has a limit $\mathbf{U}$ as $t$ approaches $a$ if its component functions approach the corresponding components of $\mathbf{U}$. For example, if $\mathbf{u}(t)=(u_1(t),u_2(t),u_3(t))$ and $\mathbf{U}=(U_1,U_2,U_3)$, then $\lim_{t\to a}\mathbf{u}(t)=\mathbf{U}$ means that $\lim_{t\to a}u_i(t)=U_i$ for each component $i$.

All the functions in a limit law must be defined for parameter values sufficiently close to $a$, apart from possibly at $a$ itself. For a sum or difference, the vector functions must be defined on a common such neighborhood. The values of the functions at $a$ do not determine their limits there.

Definition: A finite vector limit is a vector $\mathbf{U}$ such that each component of the function approaches the corresponding real component of $\mathbf{U}$. A finite scalar limit is a real number approached by the scalar function.

The Vector Limit Laws

Suppose $\lim_{t\to a}\mathbf{u}(t)=\mathbf{U}$ and $\lim_{t\to a}\mathbf{v}(t)=\mathbf{V}$, where both limits are finite vectors in the same vector space. Their sum and difference then have limits obtained by adding or subtracting the limiting vectors. If $c$ is a fixed real number, the constant multiple $c\mathbf{u}(t)$ has limit $c\mathbf{U}$.

There is also a scalar-factor law. Let $g$ be a real-valued function, with finite limit $L$, and let $\mathbf{u}$ have finite vector limit $\mathbf{U}$. Then the product $g(t)\mathbf{u}(t)$, which multiplies every component of $\mathbf{u}(t)$ by the same scalar, approaches $L\mathbf{U}$.

Theorem: If $\lim_{t\to a}\mathbf{u}(t)=\mathbf{U}$ and $\lim_{t\to a}\mathbf{v}(t)=\mathbf{V}$, then $$ \lim_{t\to a}\bigl(\mathbf{u}(t)+\mathbf{v}(t)\bigr)=\mathbf{U}+\mathbf{V}, \qquad \lim_{t\to a}\bigl(\mathbf{u}(t)-\mathbf{v}(t)\bigr)=\mathbf{U}-\mathbf{V}. $$ If $c$ is a fixed real number, then $$ \lim_{t\to a}c\mathbf{u}(t)=c\mathbf{U}. $$ If $\lim_{t\to a}g(t)=L$ for a real-valued function $g$, then $$ \lim_{t\to a}\bigl(g(t)\mathbf{u}(t)\bigr)=L\mathbf{U}. $$ In the last rule, both $L$ and $\mathbf{U}$ must be finite.

The constant-multiple rule is a special case of the scalar-factor rule: take $g(t)=c$ for every $t$. Its limit is then $c$. The theorem does not require that $\mathbf{u}(a)$ or $g(a)$ be defined.

Why the Laws Work

The reason is that vector limits are determined component by component. The ordinary scalar limit laws say that sums and differences of scalar functions with finite limits approach the sums and differences of those limits. They also say that the product of two scalar functions with finite limits approaches the product of those limits.

Here is the componentwise justification for all four rules. Write $\mathbf{u}(t)=(u_1(t),\ldots,u_n(t))$, $\mathbf{v}(t)=(v_1(t),\ldots,v_n(t))$, $\mathbf{U}=(U_1,\ldots,U_n)$, and $\mathbf{V}=(V_1,\ldots,V_n)$. The assumed vector limits give $\lim_{t\to a}u_i(t)=U_i$ and $\lim_{t\to a}v_i(t)=V_i$ for every component $i$.

Apply the scalar sum and difference laws.
For each $i$, the finite component limits imply $\lim_{t\to a}(u_i(t)+v_i(t))=U_i+V_i$ and $\lim_{t\to a}(u_i(t)-v_i(t))=U_i-V_i$. These are precisely the components of $\mathbf{U}+\mathbf{V}$ and $\mathbf{U}-\mathbf{V}$.

Apply the scalar multiple law.
For fixed $c$, each component satisfies $\lim_{t\to a}c\,u_i(t)=cU_i$, giving the components of $c\mathbf{U}$.

Apply the scalar product law.
If $\lim_{t\to a}g(t)=L$, then for each $i$, $\lim_{t\to a}g(t)u_i(t)=LU_i$, because both scalar limits are finite. Those limits are the components of $L\mathbf{U}$.

In each case, every component of the function has the corresponding component limit stated in the theorem. By the component characterization of vector limits, the vector limit therefore exists and equals the claimed vector. This also explains why finiteness is required for the scalar-factor rule: the scalar product law being used requires finite scalar limits.

Worked Examples

Worked Example: Adding and Subtracting Vector Limits

Suppose, as $t\to 0$, the component limits of $\mathbf{u}(t)$ are $1,0,0$, and those of $\mathbf{v}(t)$ are $0,1,1$. Find the limits of their sum and difference.

First, assemble the component limits into vectors: $$ \lim_{t\to 0}\mathbf{u}(t)=(1,0,0), \qquad \lim_{t\to 0}\mathbf{v}(t)=(0,1,1). $$ Both limits are finite, so the sum and difference laws apply. For the sum, add corresponding components: $$ (1,0,0)+(0,1,1)=(1+0,\,0+1,\,0+1)=(1,1,1). $$ For the difference, subtract corresponding components: $$ (1,0,0)-(0,1,1)=(1-0,\,0-1,\,0-1)=(1,-1,-1). $$ Thus the sum limit is $(1,1,1)$, and the difference limit is $(1,-1,-1)$. As a check, the first three component limits of the sum are $1+0=1$, $0+1=1$, and $0+1=1$; those of the difference are $1-0=1$, $0-1=-1$, and $0-1=-1$. They match the stated vectors.

Worked Example: Multiplying a Vector Function by a Constant

Let $\mathbf{r}(t)=(\sin t,\cos t,t)$. Find $\lim_{t\to \pi/2}(-2)\mathbf{r}(t)$.

The familiar scalar limits give $$ \lim_{t\to \pi/2}\sin t=1,\qquad \lim_{t\to \pi/2}\cos t=0,\qquad \lim_{t\to \pi/2}t=\frac{\pi}{2}. $$ Therefore $\lim_{t\to\pi/2}\mathbf{r}(t)=(1,0,\pi/2)$, a finite vector. Apply the constant-multiple law with $c=-2$: $$ \lim_{t\to\pi/2}(-2)\mathbf{r}(t) =(-2)(1,0,\pi/2) =(-2,0,-\pi). $$ The component check gives $(-2)\sin t\to-2$, $(-2)\cos t\to0$, and $(-2)t\to-\pi$. These are exactly the three components of the answer.

Worked Example: Multiplying by a Scalar Function

Suppose that as $t\to0$, $$ \lim_{t\to0}g(t)=\frac12, \qquad \lim_{t\to0}\mathbf{r}(t)=(2,1,1). $$ Find the limit of $g(t)\mathbf{r}(t)$.

The scalar limit and all components of the vector limit are finite, so the scalar-factor law applies. Multiply the scalar limit into each component: $$ \lim_{t\to0}g(t)\mathbf{r}(t) =\frac12(2,1,1) =\left(\frac12\cdot2,\,\frac12\cdot1,\,\frac12\cdot1\right) =\left(1,\frac12,\frac12\right). $$ To check, the first component product has limit $(1/2)(2)=1$, the second has limit $(1/2)(1)=1/2$, and the third has limit $(1/2)(1)=1/2$. These component limits give the stated vector.

Conditions and a Common Mistake

For the sum and difference laws, both vector limits must exist and be finite. For a constant multiple, the vector limit must exist and be finite; the constant itself is a fixed real number. For a scalar factor, both the scalar limit and the vector limit must exist and be finite. In every case, the functions must also be defined together for parameter values near the point in question.

These conditions are sufficient conditions for the conclusions. They are not claims that a limit must fail whenever one condition is missing. A product can have a limit even when its scalar factor does not have a finite limit; in that situation, this theorem simply does not establish the product limit.

Worked Example: A Product Limit When the Law Does Not Apply

Consider $g(t)=1/t$ and $\mathbf{r}(t)=(t,0,0)$ as $t\to0$, with $t\ne0$. A mistaken use of the scalar-factor law might claim $$ \lim_{t\to0}g(t)\mathbf{r}(t) =\left(\lim_{t\to0}\frac1t\right)(0,0,0) =0(0,0,0) =(0,0,0). $$ This calculation is invalid: $1/t$ has no finite limit as $t\to0$, so the scalar-factor law’s hypotheses are not satisfied.

Instead, compute the product function for every $t\ne0$: $$ g(t)\mathbf{r}(t) =\frac1t(t,0,0) =\left(\frac{t}{t},\frac0t,\frac0t\right) =(1,0,0). $$ This vector is constant on the punctured domain, so $$ \lim_{t\to0}g(t)\mathbf{r}(t)=(1,0,0). $$ The direct component check is that the first component is always $1$, while the other two are always $0$. This example shows both why the missing hypothesis matters and why its absence does not, by itself, determine whether the product has a limit.

A related error is to treat a vector limit as though the components could be combined in an arbitrary way. For example, the sum law adds first component to first component, second to second, and so on. It does not add the coordinates of one limit together or multiply unrelated components. The operation on the limiting vectors must match the operation performed on the functions.

Using the Laws Carefully

When applying a vector limit law, first identify the operation on the functions. Then verify that each required limit exists as a finite number or vector, and that the functions share a domain near the parameter value. Only after those checks should you perform the operation on the limiting values.

The laws are useful because they reduce a vector calculation to familiar scalar limit calculations. Their componentwise justification also means they apply to vectors with any fixed finite number of components, not only to vectors in three-dimensional space. They do not, however, remove the need to establish the component limits or the hypotheses of the scalar laws being used.

Key takeaway: Limits preserve sums and differences of vector functions, constant multiples of vector functions, and products of scalar and vector functions when the required finite limits exist. The resulting vector is found by applying the same operation to the limits.

Check Your Understanding

Answer each question using the conditions as well as the calculation.

If $\lim_{t\to a}\mathbf{u}(t)=\mathbf{U}$ and $\lim_{t\to a}\mathbf{v}(t)=\mathbf{V}$, what is the limit of $\mathbf{u}(t)-\mathbf{v}(t)$, and what must be true for the rule to apply?
If a vector function has a finite limit $\mathbf{U}$, what is the limit of $c\mathbf{u}(t)$ for a fixed real number $c$?
For the scalar-factor law, which two finite limits must exist?
Why can the componentwise scalar product law be used to justify the scalar-factor law for vector functions?
If a scalar factor lacks a finite limit, does that alone prove its product with a vector function has no limit? Explain.