What It Means for the Input to Grow Without Bound
A limit at infinity describes what happens to a function's values when its input moves farther and farther to the right or left. The input does not approach a real number: \(x\to+\infty\) means that \(x\) becomes arbitrarily large and positive, while \(x\to-\infty\) means that \(x\) becomes arbitrarily large in magnitude and negative.
The function's values may approach a finite number, grow without bound in one direction, or fail to settle into either kind of behavior. These possibilities are described by precise definitions. In each definition, the function must be defined for all sufficiently large positive inputs, or all sufficiently large negative inputs, as appropriate. Its behavior on a bounded part of its domain does not determine a limit at infinity.
The condition says that however small a tolerance \(\varepsilon\) is chosen around \(L\), every sufficiently large input produces a function value within that tolerance. The cutoff \(M\) can depend on \(\varepsilon\). For a limit as \(x\to-\infty\), the corresponding condition is that for every \(\varepsilon>0\), some \(M\) makes \(|f(x)-L|<\varepsilon\) whenever \(x<M\) is in the domain.
The threshold \(A\) can be as large as desired. A limit of \(+\infty\) therefore means that the function eventually exceeds every positive threshold; a limit of \(-\infty\) means it eventually falls below every negative threshold. Neither \(+\infty\) nor \(-\infty\) is a real-number value of the function or an ordinary finite limit.
Finite Limits: A Worked Example
Worked Example: A Finite Limit in Both Directions
Find and verify the limits of \(f(x)=5-\frac{6}{x}\) as \(x\to+\infty\) and \(x\to-\infty\). The function is defined for every \(x\ne 0\), so it is defined on both sufficiently distant tails.
For either direction, the distance from \(5\) is
Let \(\varepsilon>0\) be arbitrary. Choose \(M=\frac{6}{\varepsilon}\). If \(x>M\), then \(|x|=x>\frac{6}{\varepsilon}\), and hence
This verifies \(\lim_{x\to+\infty}f(x)=5\). For the negative direction, if \(x<-M\), then \(|x|=-x>M=\frac{6}{\varepsilon}\). The same calculation gives \(|f(x)-5|<\varepsilon\), so \(\lim_{x\to-\infty}f(x)=5\) as well. Thus the chosen cutoffs meet the definition in both directions; the function values approach \(5\) from below on the positive tail and from above on the negative tail.
For a numerical check, take \(\varepsilon=0.1\), so \(M=60\). At \(x=100\), \(f(100)=5-\frac{6}{100}=4.94\), and \(|4.94-5|=0.06<0.1\). At \(x=-100\), \(f(-100)=5+\frac{6}{100}=5.06\), and \(|5.06-5|=0.06<0.1\). These checks illustrate the inequality; the arbitrary-\(\varepsilon\) argument establishes the limits.
Infinite Limits: A Worked Example
Worked Example: Positive and Negative Infinite Limits
Determine the behavior of \(x^2\) and \(-x^2\) as \(x\to+\infty\) and as \(x\to-\infty\). Let \(A>0\) be any threshold, and choose \(M=\sqrt{A}\).
If \(x>M\), then \(x>\sqrt{A}\), so squaring positive quantities gives \(x^2>A\). If \(x<-M\), then \(|x|=-x>\sqrt{A}\), and therefore \(x^2=|x|^2>A\). Thus \(x^2\) exceeds every positive threshold on either tail:
For the negative of this function, the same inequalities give \(-x^2<-A\) whenever \(x>M\) or \(x<-M\). Hence
The verification is the threshold definition itself: for every \(A>0\), the cutoff \(\sqrt A\) makes the required inequality hold on the relevant tail. For instance, with \(A=9\), \(M=3\); at \(x=4\) and \(x=-4\), \(x^2=16>9\), while \(-x^2=-16<-9\). The direction of the input does not change the sign of a square.
Why the Two Input Directions Are Related
A useful change of variable turns a limit at negative infinity into one at positive infinity. Set \(t=-x\). As \(x\to-\infty\), \(t\to+\infty\), and \(f(x)=f(-t)\). More precisely, if \(g(t)=f(-t)\), then the finite limit of \(f(x)\) as \(x\to-\infty\) equals the finite limit of \(g(t)\) as \(t\to+\infty\), whenever either limit exists.
This is exactly the definition of \(\lim_{x\to-\infty}f(x)=L\). Conversely, if the negative-infinity limit of \(f\) is \(L\), take the cutoff from its definition and set \(x=-t\); the condition \(t>M\) then gives \(x<-M\), yielding the positive-infinity definition for \(g\). The same change of variable works for infinite limits because it changes the input condition but not the function value or its threshold inequality.
When No Limit Exists
A limit need not be finite or infinite. For example, a function may keep oscillating instead of settling near one number or eventually exceeding every threshold in a fixed direction. The next example verifies all three possibilities.
Worked Example: An Oscillating Function Has No Limit
Consider \(f(x)=\sin x\). We show that it has no finite limit and neither infinite limit as \(x\to+\infty\) or \(x\to-\infty\).
For the positive direction, take \(x_n=\frac{\pi}{2}+2\pi n\) and \(y_n=\frac{3\pi}{2}+2\pi n\), where \(n\) is a positive integer. Both sequences of inputs tend to \(+\infty\), while periodicity gives
If \(\sin x\) had a finite limit \(L\) at \(+\infty\), the definition would force both sets of function values to approach \(L\). But the first set is constantly \(1\) and the second constantly \(-1\). A constant sequence with value \(1\) cannot approach a different number \(L\): if \(L\ne1\), choosing \(\varepsilon=\frac{|1-L|}{2}\) contradicts \(|1-L|<\varepsilon\). The same argument applies to the constant value \(-1\). Thus a finite limit would have to equal both \(1\) and \(-1\), which is impossible.
For \(x\to-\infty\), use \(u_n=\frac{\pi}{2}-2\pi n\) and \(v_n=\frac{3\pi}{2}-2\pi n\). Both tend to \(-\infty\), and their sine values are \(1\) and \(-1\), respectively. The identical contradiction rules out a finite limit in this direction too.
Finally, \(-1\le\sin x\le1\) for all real \(x\). A limit of \(+\infty\) would require \(\sin x>2\) for every sufficiently distant input, and a limit of \(-\infty\) would require \(\sin x<-2\). Both requirements contradict the range bound, in either direction. Therefore \(\sin x\) has no finite or infinite limit as the input tends to either infinity.
Reading the Definitions Correctly
The definitions distinguish the direction of the input from the behavior of the output. The notation \(x\to-\infty\) says that inputs are negative and increasingly large in magnitude; it does not say that the function values must become negative. For example, the worked example shows that \(x^2\to+\infty\) in both directions.
A common error is to infer the output limit directly from the sign of the input. For instance, the calculation “\(x\to-\infty\), so \(x^2\to-\infty\)” is wrong. When \(x<-\sqrt A\), we have \(|x|>\sqrt A\), so \(x^2=|x|^2>A\), not \(x^2<-A\). The corrected conclusion is \(x^2\to+\infty\) as \(x\to-\infty\).
Another important distinction is that \(x\to+\infty\) does not imply \(f(x)\to+\infty\). The definition concerns the input only; the output must be examined separately. The finite-limit example, for instance, has arbitrarily large positive inputs but function values approach \(5\). To decide what happens, ask which precise condition applies:
| Behavior sought | Condition on distant inputs |
|---|---|
| Finite limit \(L\) | Function values eventually lie within every positive tolerance of \(L\). |
| Limit \(+\infty\) | Function values eventually exceed every positive threshold. |
| Limit \(-\infty\) | Function values eventually fall below every negative threshold. |
| No limit | None of those conditions holds. |
These definitions also clarify why checking a few inputs cannot establish a limit: the requirement concerns every input sufficiently far along the chosen tail. A useful proof identifies a cutoff depending on the requested tolerance or threshold, then verifies the required inequality for all inputs beyond that cutoff.
Summary
A finite limit at infinity means that the function values can be forced arbitrarily close to a real number by taking the input sufficiently far in the specified direction. An infinite limit means that the function values eventually pass every threshold in the specified direction. Oscillation or other behavior may prevent either type of limit from existing. Always keep the two questions separate: which way is the input moving, and what do the function values do there?
Check Your Understanding
For each question, use the definitions and distinguish the direction of the input from the behavior of the function values.
- State the definition of a finite limit \(L\) as \(x\to-\infty\).
- Why does \(x^2\to+\infty\) as \(x\to-\infty\), rather than \(-\infty\)?
- What must be shown, for an arbitrary \(A>0\), to verify that \(f(x)\to-\infty\) as \(x\to+\infty\)?
- Why do the alternating values of \(\sin x\) along inputs tending to \(+\infty\) rule out a finite limit?
- How does the substitution \(t=-x\) relate a limit as \(x\to-\infty\) to a limit as \(t\to+\infty\)?
