Limits at Infinite Parameter Values

What a Limit at Infinity Describes

A vector-valued function \(\mathbf{r}(t)\) describes a vector for each allowed parameter value \(t\). Asking for its limit as \(t\) tends to positive infinity means asking whether those vectors approach one fixed, finite vector as \(t\) becomes arbitrarily large. A limit at negative infinity asks the same question as \(t\) decreases without bound.

These are limits of a function’s values, not values obtained by substituting infinity for \(t\). Infinity is not a real number that can be substituted into a formula. Instead, the behavior of each coordinate is analyzed as the parameter grows in magnitude.

Definition: For a vector-valued function with components \(f_1,\ldots,f_n\), the finite limit \(\lim_{t\to\infty}\mathbf{r}(t)=\mathbf{L}\) means that \(\mathbf{r}(t)\) approaches the fixed vector \(\mathbf{L}\) as \(t\) becomes arbitrarily large. The definition of \(\lim_{t\to-\infty}\mathbf{r}(t)=\mathbf{L}\) uses \(t\) becoming arbitrarily negative. In both cases, the limit is found from the component limits:

\mathbf{r}(t)=\langle f_1(t),\ldots,f_n(t)\rangle, \qquad \mathbf{L}=\langle L_1,\ldots,L_n\rangle, \qquad \lim_{t\to\pm\infty}\mathbf{r}(t) = \left\langle \lim_{t\to\pm\infty}f_1(t),\ldots, \lim_{t\to\pm\infty}f_n(t) \right\rangle,

provided each component limit exists as a finite real number. The componentwise rule for vector limits applies here just as it does at a finite parameter value. If a component grows without bound, the vector cannot approach a finite vector: the distance between the vectors is at least the absolute difference in that coordinate.

Evaluating Polynomial Components

A nonconstant polynomial does not approach a finite value as \(t\) tends to either positive or negative infinity. Its highest-degree term determines its unbounded behavior. For example, \(3t^2\) grows positively at both ends, whereas \(-2t\) grows positively as \(t\to-\infty\) and negatively as \(t\to\infty\). A constant polynomial, by contrast, has the same finite value at both ends.

Worked Example: A Polynomial Vector with No Finite Limit

Consider \(\mathbf{p}(t)=\langle 3t^2-1,-2t+4,7\rangle\). Determine whether it has a finite limit as \(t\to-\infty\).

Evaluate the first component. Since \(t^2\to+\infty\), we have \(3t^2\to+\infty\). Subtracting \(1\) does not change this behavior:

3t^2-1\longrightarrow+\infty.

For the second component, \(-2t\to+\infty\) as \(t\to-\infty\), so adding \(4\) still gives an unbounded component:

-2t+4\longrightarrow+\infty.

The third component is constant, and therefore \(7\to7\). The first and second components do not have finite limits. In particular, a finite vector limit is impossible. To see this directly, if a proposed limit had first coordinate \(L_1\), then the distance from \(\mathbf{p}(t)\) to that vector would be at least \(\lvert 3t^2-1-L_1\rvert\), which grows without bound.

Thus \(\mathbf{p}(t)\) has no finite vector limit as \(t\to-\infty\). The component calculations provide the check: two coordinates diverge, even though the third remains constant.

Rational Functions: Compare Degrees

For a rational component \(P(t)/Q(t)\), where \(P\) and \(Q\) are polynomials and \(Q(t)\ne0\) for sufficiently large \(\lvert t\rvert\), compare their degrees. This works because the highest-degree terms control the quotient at large parameter values.

Suppose \(P\) has degree \(m\) and leading coefficient \(a\), while \(Q\) has degree \(n\) and leading coefficient \(b\ne0\). If the degrees are equal, the quotient tends to \(a/b\). If the numerator has lower degree, the quotient tends to \(0\). If the numerator has higher degree, its magnitude grows without bound, so there is no finite limit. When considering \(t\to-\infty\), also track the sign of the remaining power of \(t\).

Why the degree comparison works: Divide the numerator and denominator by \(t^n\), where \(n\) is the denominator’s degree. Terms with negative powers of \(t\) tend to \(0\) at either end. If \(m=n\), the remaining leading terms give \(a/b\); if \(m<n\), the numerator tends to \(0\) while the denominator tends to \(b\). If \(m>n\), the quotient retains a positive power of \(t\), multiplied by a nonzero leading-coefficient ratio, and cannot approach a finite value.

This gives a useful test for finite vector limits: apply the degree comparison to every rational component, and then assemble the finite component limits into a vector. If even one component has numerator degree greater than denominator degree, the vector has no finite limit.

Worked Example: Rational Components with Different Degrees

Find \(\lim_{t\to\infty}\mathbf{r}(t)\) for

\mathbf{r}(t)= \left\langle \frac{6t^2+2t-1}{3t^2-4}, \frac{t+5}{t^2+1}, \frac{4t^3-2}{8t^3+t^2} \right\rangle.

For the first component, divide the numerator and denominator by \(t^2\):

\frac{6t^2+2t-1}{3t^2-4} = \frac{6+\frac{2}{t}-\frac{1}{t^2}} {3-\frac{4}{t^2}} \longrightarrow \frac{6}{3}=2.

For the second component, divide by \(t^2\), the denominator’s highest power:

\frac{t+5}{t^2+1} = \frac{\frac{1}{t}+\frac{5}{t^2}} {1+\frac{1}{t^2}} \longrightarrow \frac{0+0}{1+0}=0.

For the third component, divide numerator and denominator by \(t^3\):

\frac{4t^3-2}{8t^3+t^2} = \frac{4-\frac{2}{t^3}} {8+\frac{1}{t}} \longrightarrow \frac{4}{8}=\frac{1}{2}.

Each component has a finite limit, so the vector limit is

\lim_{t\to\infty}\mathbf{r}(t) = \left\langle 2,0,\frac{1}{2}\right\rangle.

As a check, the three coordinate limits, in order, are \(2\), \(0\), and \(1/2\); these are exactly the entries of the stated limit vector.

Negative Infinity and Signs

The degree comparison also applies as \(t\to-\infty\), since \(1/t^k\to0\) for every positive integer \(k\). However, a power of \(t\) may be negative at that end, so do not infer the sign from the degree alone. For instance, \(t^2\) is positive at both ends, but \(t\) is negative as \(t\to-\infty\).

Worked Example: A Finite Limit at Negative Infinity

Find the limit of

\mathbf{s}(t)= \left\langle \frac{5t^2-3t+2}{2t^2+t}, \frac{3t-1}{t^2+4}, \frac{7t^3+t}{-2t^3+5} \right\rangle

as \(t\to-\infty\). Divide the first numerator and denominator by \(t^2\):

\frac{5t^2-3t+2}{2t^2+t} = \frac{5-\frac{3}{t}+\frac{2}{t^2}} {2+\frac{1}{t}} \longrightarrow \frac{5}{2}.

For the second component, divide by \(t^2\):

\frac{3t-1}{t^2+4} = \frac{\frac{3}{t}-\frac{1}{t^2}} {1+\frac{4}{t^2}} \longrightarrow \frac{0-0}{1+0}=0.

For the third component, divide by \(t^3\), keeping the signs in the denominator:

\frac{7t^3+t}{-2t^3+5} = \frac{7+\frac{1}{t^2}} {-2+\frac{5}{t^3}} \longrightarrow \frac{7}{-2}=-\frac{7}{2}.

All three limits are finite, so

\lim_{t\to-\infty}\mathbf{s}(t) = \left\langle\frac{5}{2},0,-\frac{7}{2}\right\rangle.

The check is componentwise: the limits just computed match the first, second, and third coordinates of the result. Notice that dividing by \(t^3\) in the last component leaves the leading denominator coefficient \(-2\), not \(2\).

A Pitfall: Losing the Sign at Negative Infinity

A common error is to treat negative infinity as though the parameter were positive and large. For example, the quotient \(t^2/t\) is not positive and large when \(t\) tends to negative infinity. Simplifying first makes its sign clear.

\frac{t^2}{t}=t\longrightarrow-\infty \qquad (t\to-\infty).

By contrast, \(t^2\to+\infty\). These two expressions have different signs at negative infinity because the first retains one factor of \(t\). When simplifying a rational component, keep track of any remaining odd power of \(t\); it changes sign between the two ends of the real line.

A Reliable Procedure

For a vector limit at infinity, work through the components in order and do not assemble the result until each component has been checked.

Write the components separately.
Identify each scalar function in the vector, including any polynomial or rational expression.

Evaluate each scalar limit.
For rational functions, compare degrees or divide by a suitable power of \(t\). For polynomials, use the highest-degree term.

Check signs at the chosen end.
When \(t\to-\infty\), account for whether the remaining power of \(t\) is odd or even.

Assemble only finite limits.
If every component has a finite limit, place those values in order to form the limit vector. If a component diverges, there is no finite vector limit.

This method matters because a vector limit is a statement about all of the vector’s coordinates, not just its overall appearance or one especially simple component. Checking each coordinate prevents a divergent component from being overlooked and makes it clear whether the limiting vector exists.

Check Your Understanding

For each question, consider the behavior of every component and pay attention to whether the parameter tends to positive or negative infinity.

What is the limit of \(\langle 4,\,2t/(t^2+1)\rangle\) as \(t\to\infty\)?
Does \(\langle t^2+1,\,3\rangle\) have a finite vector limit as \(t\to-\infty\)? Explain.
What is the limit of \(\left\langle (3t^2+1)/(t^2-2),\,(t-4)/(t^2+1)\right\rangle\) as \(t\to-\infty\)?
Why does a component whose magnitude grows without bound prevent a finite vector limit?
As \(t\to-\infty\), does \(t^3/t^2\) tend to positive infinity or negative infinity?