How to Calculate the Limit of a Vector

To calculate the limit of a vector-valued function, take the limit of each component separately. The vector limit exists when every component has a finite limit, and its components are those scalar limits.

This rule applies to limits of curves such as $\mathbf{r}(t)$ as $t$ approaches a number. For a function of several variables, the same component-by-component rule applies, but each scalar component must have a genuine limit independent of the path of approach.

The componentwise rule

Suppose a vector function takes values in $\mathbb{R}^n$:

$$\mathbf{f}(t)=\bigl(f_1(t),f_2(t),\ldots,f_n(t)\bigr).$$

Then

$$\lim_{t\to a}\mathbf{f}(t)=\mathbf{L}$$

if and only if each scalar limit $\lim_{t\to a} f_i(t)$ exists and is finite. In that case,

$$\lim_{t\to a}\mathbf{f}(t)=\left(\lim_{t\to a}f_1(t),\lim_{t\to a}f_2(t),\ldots,\lim_{t\to a}f_n(t)\right).$$

This is not just a convenient computational shortcut. It follows from the usual definition of a vector limit using distance: the distance between $\mathbf{f}(t)$ and $\mathbf{L}$ must approach zero. In finite-dimensional Euclidean space, that happens exactly when the difference in every coordinate approaches zero.

Worked example

Evaluate

$$\lim_{t\to 0}\left(\frac{\sin(3t)}{t},\frac{1-\cos t}{t^2},\frac{e^t-1}{t}\right).$$

Find the limit of each component:

$$\lim_{t\to 0}\frac{\sin(3t)}{t} =3\lim_{t\to 0}\frac{\sin(3t)}{3t}=3,$$

using the standard limit $\lim_{u\to 0}\sin u/u=1$.

For the second component, use $1-\cos t=2\sin^2(t/2)$:

$$\lim_{t\to 0}\frac{1-\cos t}{t^2} =\frac{1}{2}\lim_{t\to 0}\left(\frac{\sin(t/2)}{t/2}\right)^2 =\frac{1}{2}.$$

The third component is the standard exponential limit:

$$\lim_{t\to 0}\frac{e^t-1}{t}=1.$$

All three component limits exist and are finite, so reassemble them in the original order:

$$\lim_{t\to 0}\left(\frac{\sin(3t)}{t},\frac{1-\cos t}{t^2},\frac{e^t-1}{t}\right) =\left(3,\frac{1}{2},1\right).$$

If a component does not have a limit

A finite vector limit cannot exist if even one component fails to approach a finite value. For example,

$$\mathbf{g}(t)=\left(t,\frac{1}{t}\right)$$

has no finite limit as $t\to 0$, because its second component is unbounded. The first component approaching zero does not make up for that failure.

A component can also fail to have a limit by oscillating. For instance, $(t,\sin(1/t))$ has no vector limit as $t\to 0$: the second coordinate does not settle to one number.

Limits of functions of several variables

For a function $\mathbf{F}(x,y)$, find the limit of each scalar component as $(x,y)$ approaches the target point. Each component must have the same limit along every path to that point. Checking a few paths can show that a limit does not exist if they give different answers, but matching answers along a few paths does not prove that the limit exists. Use a bound or a definition-based argument when path checks are inconclusive.

For example, if $\mathbf{F}(x,y)=(F_1(x,y),F_2(x,y))$, then a limit at $(a,b)$ exists exactly when both $F_1$ and $F_2$ have limits there; the vector limit consists of those two limits.

Common mistakes

• Taking the limit of the vector as if it were one scalar expression. Split it into components first.
• Ignoring a bad component. Every component must have a finite limit for an ordinary vector limit to exist.
• Using different approaches for different components. All component limits concern the same input approaching the same point.
• Assuming a few paths prove a multivariable limit. Different path results disprove a limit, but agreement along selected paths alone is not a proof.
• Confusing a vector limit with a limit of its length. The length $\|\mathbf{f}(t)\|$ may have a limit even when the vector itself does not. For example, $(\cos(1/t),\sin(1/t))$ has constant length $1$ but no vector limit as $t\to 0$.