When Bounds Leave Too Much Room
Bounds can be powerful tools for evaluating limits. If a function is trapped between two functions that approach the same number, the trapped function must approach that number too. But if the bounds approach different numbers, the interval between them does not close to a single value. It may still be possible to find the enclosed function’s limit by using additional information, but these bounds alone do not determine it.
This distinction matters especially with oscillatory functions. A function may keep moving between values inside its bounds, or its oscillations may become smaller even though the available bounds stay wide. In both cases, the limiting values of the bounds are different, so the squeeze theorem does not settle the question.
The Squeeze Theorem and Its Condition
Suppose \(f\), \(g\), and \(h\) are defined for \(x\) near \(a\), possibly excluding \(a\), and that \(g(x)\leq f(x)\leq h(x)\) throughout that punctured neighborhood. The squeeze theorem says that if \(g\) and \(h\) both approach the same finite number \(L\) as \(x\) approaches \(a\), then \(f\) also approaches \(L\).
The matching limit is essential. Here is why. Given any \(\varepsilon>0\), the two bounds eventually satisfy \(L-\varepsilon<g(x)\) and \(h(x)<L+\varepsilon\). Since \(g(x)\leq f(x)\leq h(x)\), these inequalities force \(L-\varepsilon<f(x)<L+\varepsilon\). That is exactly what it means for \(f(x)\) to approach \(L\).
If the lower bound instead approaches \(A\) and the upper bound approaches \(B\), where \(A<B\), that reasoning no longer traps \(f(x)\) in an arbitrarily small interval around one number. The limiting interval has width \(B-A>0\). The function might converge to a value inside that interval, or it might fail to converge.
Worked Example: An Oscillation Trapped Between Different Limits
Worked Example: An Oscillation Trapped Between Different Limits
Determine whether \(f(x)=\sin(1/x)\) has a limit as \(x\to0^+\), using the bounds \(-1\leq f(x)\leq1\).
The bounds hold because the sine function always lies between \(-1\) and \(1\). Their limits are
These are different values, so the squeeze theorem does not apply. To determine whether the function itself has a limit, choose positive inputs at which the sine takes alternating extreme values. For \(n=1,2,3,\ldots\), define
Both sequences are positive and tend to \(0\), because their denominators grow without bound. Substitution gives
If \(f(x)\) had a limit as \(x\to0^+\), its values along every sequence of positive inputs tending to \(0\) would approach that same limit. Along \(x_n\), the values are constantly \(1\); along \(y_n\), they are constantly \(-1\). They do not approach the same number. Therefore, \(\lim_{x\to0^+}\sin(1/x)\) does not exist. The bounds are valid, but their different limiting values neither establish nor rule out a limit; the sequences settle this example.
Wide Bounds Can Hide Convergence Too
Different limiting values for the bounds do not prove that the enclosed function fails to have a limit. They only show that this particular squeeze argument is inconclusive. A function can converge while remaining between bounds that leave plenty of room.
Worked Example: A Convergent Oscillation Inside Wide Bounds
Consider \(F(x)=1+x\sin(1/x)\) as \(x\to0^+\). First use the range of sine, \(-1\leq\sin(1/x)\leq1\). Since \(x>0\), multiplication by \(x\) preserves the inequalities, so
For \(0<x<1\), these imply the wider bounds \(0\leq F(x)\leq2\). The limits of these wider bounds are \(0\) and \(2\), which are different. They do not establish the limit of \(F\).
The original inequalities give a sharper estimate. Subtract \(1\) and take absolute values:
As \(x\to0^+\), the right side \(x\) approaches \(0\). Therefore \(|F(x)-1|\to0\), which means \(F(x)\to1\). The oscillatory factor does not disappear, but its amplitude is multiplied by \(x\), so the oscillations shrink. The wide bounds \(0\leq F(x)\leq2\) conceal this convergence; the sharper estimate reveals it.
For a direct check, \(F(x_n)=1+x_n\) and \(F(y_n)=1-y_n\) for the sequences from the previous example. Since \(x_n\to0\) and \(y_n\to0\), these values approach \(1\) from opposite sides, consistent with the limit just established.
The Same Bounds Can Allow Different Outcomes
A useful way to see the limitation is to keep the bounds fixed and change only the enclosed function. The bounds \(0\leq f(x)\leq2\) for \(0<x<1\) allow a function to converge, and they also allow one to oscillate without a limit.
Worked Example: One Pair of Bounds, Two Behaviors
For \(0<x<1\), the constant function \(p(x)=1\) satisfies \(0\leq p(x)\leq2\), and \(\lim_{x\to0^+}p(x)=1\).
Now let \(q(x)=1+\sin(1/x)\). Since \(-1\leq\sin(1/x)\leq1\), adding \(1\) gives
Thus \(p\) and \(q\) satisfy exactly the same bounds. But \(q(x_n)=2\) and \(q(y_n)=0\) for the sequences already defined. Because these values do not approach a common number, \(q\) has no limit as \(x\to0^+\). The bounds alone cannot distinguish the convergent function \(p\) from the nonconvergent function \(q\).
Check: \(p(x)=1\) is always inside the interval \([0,2]\), and \(q(x)\) is inside it because the sine values lie in \([-1,1]\). The sequence values for \(q\) are exactly \(2\) and \(0\), confirming its oscillation does not settle.
What Different Limiting Bounds Can Tell You
Although different limiting values do not determine whether the enclosed function converges, they can constrain its possible limit if a limit is known to exist. Suppose \(g(x)\leq f(x)\leq h(x)\), and the limits of \(g\), \(f\), and \(h\) exist as \(x\to a\), with values \(A\), \(L\), and \(B\), respectively. Then \(A\leq L\leq B\).
For example, if \(L>A\), choose \(\varepsilon=(L-A)/3\), which is positive. Near \(a\), convergence gives \(g(x)<A+\varepsilon\) and \(f(x)>L-\varepsilon\). Since \(A+\varepsilon<L-\varepsilon\), the order \(g(x)\leq f(x)\) is consistent with the limits. The reverse order \(L<A\) would produce the opposite contradiction: choosing \(\varepsilon=(A-L)/3\) would make \(f(x)<L+\varepsilon<A-\varepsilon<g(x)\) sufficiently near \(a\), contrary to \(g(x)\leq f(x)\). Hence \(A\leq L\). Applying the same reasoning to \(f(x)\leq h(x)\) gives \(L\leq B\).
A Common Mistake: Confusing Boundedness with Convergence
A common incorrect argument is: “Since \(-1\leq\sin(1/x)\leq1\), the function is bounded, so it must have a limit.” The conclusion does not follow. Boundedness says the function’s values stay within a fixed interval; convergence requires those values to get arbitrarily close to one number.
The incorrect argument applied to \(\sin(1/x)\) would claim that its limit exists just because it remains between \(-1\) and \(1\). But the sequences \(x_n\to0^+\) and \(y_n\to0^+\) give \(f(x_n)=1\) and \(f(y_n)=-1\), so the function has no limit. The correction is to check whether the bounds approach the same value—or, if they do not, to obtain a sharper estimate or use another method.
How to Diagnose a Wide Squeeze
When an inequality traps a function between bounds with different limits, do not force the squeeze theorem to give a conclusion it does not support. First identify what the bounds actually establish: the function stays between them near the point. Then ask whether the interval between the bounds shrinks to a point. If it does not, consider whether a sharper inequality is available. For an oscillatory expression, it is often useful to separate the oscillating factor from a factor that may approach zero.
If they differ, the squeeze theorem does not establish the enclosed limit.
The enclosed function may still converge, as \(1+x\sin(1/x)\) does.
Try a sharper estimate, or test oscillation with sequences approaching the input value.
The central point is logical: bounds with different limits are not useless, but they leave a gap. A shrinking gap forces a unique limiting value; a gap that remains open permits multiple possibilities. To decide what happens, you need information beyond those wide bounds.
Check Your Understanding
For each question, distinguish what the bounds prove from what they leave undecided.
- If \(g(x)\leq f(x)\leq h(x)\), with \(g(x)\to-2\) and \(h(x)\to3\), does the squeeze theorem establish a limit for \(f\)? Explain.
- Why do the sequences \(x_n\) and \(y_n\) in the first worked example show that \(\sin(1/x)\) has no limit as \(x\to0^+\)?
- For \(F(x)=1+x\sin(1/x)\), what sharper inequality establishes its limit, and what is that limit?
- Can two functions satisfying \(0\leq f(x)\leq2\) have different limit behavior as \(x\to0^+\)? Give the pair used above.
- If the lower and upper bounds have different limits, but the enclosed function is known to have a limit \(L\), what can the bounds tell you about \(L\)?
