One-Sided Limits at Parameter Endpoints

Why the Side of Approach Matters

A vector-valued function can be defined only for parameter values on one side of a finite endpoint. Its limit there must be taken from that accessible side. For example, if the parameter belongs to $[0,1]$, values in the domain can approach $0$ from the right, but not from the left.

At a finite endpoint, the direction of approach is part of the question. The notation $t\to a^-$ means that $t$ approaches $a$ through values less than $a$; $t\to a^+$ means that $t$ approaches through values greater than $a$. A limit from a side is meaningful only if the function's domain contains parameter values arbitrarily close to $a$ on that side.

Key idea: A one-sided vector limit describes the vector values approached from one specified side of the parameter. Use only parameter values in the function's domain, and do not substitute the opposite side.

Definition of a One-Sided Vector Limit

Let $\mathbf{r}:D\to\mathbb{R}^n$ be a vector-valued function, and let $a$ be a finite real number. For a left-hand limit, the domain must contain points less than $a$ arbitrarily close to $a$. We write $\lim_{t\to a^-}\mathbf{r}(t)=\mathbf{L}$ if the vector values approach $\mathbf{L}$ as $t$ approaches $a$ from that side.

Precisely, this means that for every $\varepsilon>0$, there is a $\delta>0$ such that, whenever $t\in D$ and $a-\delta<t<a$,

\|\mathbf{r}(t)-\mathbf{L}\|<\varepsilon.

The right-hand limit $\lim_{t\to a^+}\mathbf{r}(t)=\mathbf{L}$ is defined in the same way, except that $a<t<a+\delta$. Here $\|\cdot\|$ denotes the usual Euclidean norm. In either definition, there must be domain points arbitrarily close to $a$ from the specified side; otherwise there is no approach from that side on which to define the limit.

The limit concerns values of $\mathbf{r}(t)$ for $t$ near $a$, not necessarily the value at $a$. The function may be undefined at $a$, or it may have a value there different from the limit.

Notation: The superscript minus in $\lim_{t\to a^-}$ indicates approach from below, while the superscript plus in $\lim_{t\to a^+}$ indicates approach from above. They describe direction, not the sign of the limit vector.

Using Component Limits at an Endpoint

Write a vector-valued function in components as $\mathbf{r}(t)=(f_1(t),\ldots,f_n(t))$. The component-limit result from the earlier material on vector limits applies to one-sided limits as well: the one-sided vector limit exists and equals $(L_1,\ldots,L_n)$ exactly when each corresponding scalar one-sided limit exists and equals $L_i$. Thus, the side must be kept the same for every component.

\lim_{t\to a^\pm}\mathbf{r}(t) = \left( \lim_{t\to a^\pm}f_1(t),\ldots, \lim_{t\to a^\pm}f_n(t) \right),

provided all the component limits on the indicated side exist as finite real numbers. If even one component fails to have a finite limit, the vector cannot have a finite limit on that side. The endpoint value itself is not needed for this component-by-component evaluation.

Worked Example: A Right-Hand Limit at a Domain Endpoint

Let $\mathbf{r}(t)=\left(\sqrt{t+2},\frac{t^2-4}{t+2},\frac{t+2}{t+5}\right)$, defined for $t>-2$. Find $\lim_{t\to-2^+}\mathbf{r}(t)$.

The domain approaches $-2$ from the right, so the right-hand limit is the appropriate one. For $t>-2$, the second component can be simplified because $t+2\ne0$:

\frac{t^2-4}{t+2} = \frac{(t-2)(t+2)}{t+2} = t-2.

Now take each component limit from the right:

\lim_{t\to-2^+}\sqrt{t+2}=0,\qquad \lim_{t\to-2^+}(t-2)=-4,\qquad \lim_{t\to-2^+}\frac{t+2}{t+5}=\frac{0}{3}=0.

All three component limits are finite, so the vector limit is

\lim_{t\to-2^+}\mathbf{r}(t)=(0,-4,0).

As a check, the first component approaches $0$ as the square root's input approaches $0$ through positive values; the simplified second component approaches $-4$; and the third approaches $0/3=0$. These are exactly the entries of the stated limit vector.

Worked Example: A Left-Hand Limit at a Domain Endpoint

For $t<3$, let $\mathbf{r}(t)=\left(\frac{t^2-9}{t-3},\cos(\pi t),\sqrt{4-t}\right)$. Find $\lim_{t\to3^-}\mathbf{r}(t)$.

The domain is restricted to values below $3$, so approach from the left. For $t<3$, the denominator $t-3$ is nonzero, and the first component simplifies as follows:

\frac{t^2-9}{t-3} = \frac{(t-3)(t+3)}{t-3} = t+3.

The three component limits are

\lim_{t\to3^-}(t+3)=6,\qquad \lim_{t\to3^-}\cos(\pi t)=\cos(3\pi)=-1,\qquad \lim_{t\to3^-}\sqrt{4-t}=\sqrt{1}=1.

Consequently,

\lim_{t\to3^-}\mathbf{r}(t)=(6,-1,1).

To check, the simplified first component tends to $6$, the cosine tends to $-1$, and the square root tends to $1$. The original first component is undefined at $t=3$, but that does not prevent the left-hand limit from existing.

Why a One-Sided Limit Is Unique

If a one-sided vector limit exists, it cannot have two different values. This follows directly from the norm definition and helps explain why the result of a calculation is unambiguous.

Proof: Suppose $\lim_{t\to a^+}\mathbf{r}(t)=\mathbf{L}$ and $\lim_{t\to a^+}\mathbf{r}(t)=\mathbf{M}$. If $\mathbf{L}\ne\mathbf{M}$, set $d=\|\mathbf{L}-\mathbf{M}\|>0$. By the two limit definitions, there are positive numbers $\delta_L$ and $\delta_M$ such that, for domain values $t$ just to the right of $a$, $\|\mathbf{r}(t)-\mathbf{L}\|<d/3$ and $\|\mathbf{r}(t)-\mathbf{M}\|<d/3$, respectively. Take $t$ close enough to $a$ that both inequalities apply; such a domain value exists because the right-hand approach is available. The triangle inequality then gives $$ d=\|\mathbf{L}-\mathbf{M}\| \le \|\mathbf{L}-\mathbf{r}(t)\|+\|\mathbf{r}(t)-\mathbf{M}\| <\frac{2d}{3}, $$ which contradicts $d>0$. Therefore $\mathbf{L}=\mathbf{M}$. The proof for a left-hand limit is identical, using domain values to the left of $a$.

When a Finite One-Sided Limit Does Not Exist

A finite vector limit requires every component to approach a finite real number. A component that grows without bound is enough to rule out such a limit.

Worked Example: An Unbounded Component Prevents a Limit

Let $\mathbf{r}(t)=\left(\frac{1}{t},\frac{t}{t+1}\right)$ for $0<t\le1$. Determine whether a finite right-hand limit exists at the endpoint $t=0$.

The domain approaches $0$ from the right. The first component $1/t$ is unbounded as $t\to0^+$. More explicitly, for any proposed finite vector $\mathbf{L}=(L_1,L_2)$, choose $t>0$ small enough that $1/t>|L_1|+2$. Then

\left|\frac{1}{t}-L_1\right| \ge \frac{1}{t}-|L_1| >2.

The distance between two vectors is at least the absolute difference between any corresponding components. Therefore $\|\mathbf{r}(t)-\mathbf{L}\|>2$ for such $t$, so it cannot be made arbitrarily small by taking $t$ closer to $0$. No finite right-hand vector limit exists. As a check on the other component, $\lim_{t\to0^+}t/(t+1)=0$; that finite limit does not fix the failure of the first component.

Common Pitfall: Using the Wrong Side

Suppose $\mathbf{q}(t)=(\sqrt{t},t)$ is defined for $0\le t\le1$. A wrong approach to the endpoint $0$ would be to try $t\to0^-$ and reason that $\sqrt{t}$ is not real for negative $t$, so no limit exists. This uses values outside the domain and tests the wrong side.

The correct approach is from the right, where the domain has values arbitrarily close to $0$. Both components have right-hand limits:

\lim_{t\to0^+}\sqrt{t}=0, \qquad \lim_{t\to0^+}t=0, \qquad \text{so}\qquad \lim_{t\to0^+}\mathbf{q}(t)=(0,0).

Always identify the domain near the endpoint before calculating. An endpoint of an interval of parameters typically has only one accessible side. At an endpoint approached from below, use a left-hand limit; at an endpoint approached from above, use a right-hand limit.

Why Endpoint Limits Matter

Finite-endpoint limits describe how a parametrized curve behaves as its parameter approaches the start or end of a specified parameter interval. They can identify the limiting position even when the parametrization is not defined at the endpoint, or when its endpoint value differs from the position it approaches. They also provide a precise check that each coordinate of the curve has a finite endpoint value in the limiting sense.

For a practical calculation, first determine which side the domain permits. Then evaluate the scalar one-sided limits component by component, and assemble them into a vector only if all are finite. This order avoids treating an inaccessible side as meaningful or overlooking a component that prevents the limit from existing.

Check Your Understanding

For each question, pay attention to the parameter domain and the direction of approach.

If $\mathbf{r}:D\to\mathbb{R}^3$ has domain $D=[2,5]$, which one-sided limit is appropriate at the endpoint $2$?
For $\mathbf{r}(t)=(t^2,\sqrt{t})$, $0\le t\le1$, find the right-hand vector limit as $t\to0$.
Can a vector have a finite one-sided limit if one of its components is unbounded on that side? Explain.
Does a function need to be defined at $t=a$ for its left-hand limit there to exist?
For $\mathbf{r}(t)=(t,1/t)$, $0<t\le1$, determine whether a finite right-hand vector limit exists at $t=0$.