When Bounds Determine a Limit
A function can be difficult to evaluate directly near a point, even when its values are constrained by simpler expressions. If those constraints hold for every nearby input, and the simpler expressions approach the same number, then the function between them must approach that number too.
This is the idea behind the Squeeze Theorem. Its force comes not just from having inequalities, but from having valid inequalities throughout a neighborhood on both sides of the point.
Precise Hypotheses
Let \(a\) be a real number. A punctured neighborhood of \(a\) is an interval around \(a\) with the point \(a\) itself removed. In limit notation, we can specify one by \(0<|x-a|<\delta\), where \(\delta>0\). Excluding \(a\) matters because a limit concerns values near \(a\), not necessarily the value of a function at \(a\).
The Squeeze Theorem uses three functions \(f\), \(g\), and \(h\), all defined on some punctured neighborhood of \(a\). The inequalities must hold at every point in a sufficiently small such neighborhood, not just at selected inputs.
\(g(x)\le f(x)\le h(x)\).
If \(\lim_{x\to a}g(x)=L\) and \(\lim_{x\to a}h(x)=L\), where \(L\) is a finite real number, then \(\lim_{x\to a}f(x)=L\).
The limit requirement applies to both bounding functions, and their limits must be the same value. It is not enough for just one bound to approach \(L\), or for the two bounds to approach different values. The theorem is a sufficient condition for a limit; if its conditions are not met, that only means this theorem does not settle the limit.
Why the Theorem Works
Here is a proof using the definition of a limit. To show that \(\lim_{x\to a}f(x)=L\), we must show that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\varepsilon\).
For a given \(\varepsilon>0\), choose a neighborhood small enough that \(L-\varepsilon<g(x)\).
Shrink the neighborhood if necessary so that \(h(x)<L+\varepsilon\).
Also choose the neighborhood small enough for \(g(x)\le f(x)\le h(x)\) to hold. Then \(L-\varepsilon<f(x)<L+\varepsilon\), which means \(|f(x)-L|<\varepsilon\).
To make the choices precise, the limit of \(g\) supplies a radius \(\delta_g>0\) for the first inequality, and the limit of \(h\) supplies a radius \(\delta_h>0\) for the second. Let \(\delta\) be the smallest of these two radii and a radius \(\delta_b>0\) on which the bounds hold. For every \(x\) with \(0<|x-a|<\delta\), all three conditions hold. Thus
Subtracting \(L\) shows that \(-\varepsilon<f(x)-L<\varepsilon\), which is equivalent to \(|f(x)-L|<\varepsilon\). This works for every \(\varepsilon>0\), so the definition of the limit gives \(\lim_{x\to a}f(x)=L\).
This proof also explains why two-sided bounds are useful: they control both how far below \(L\) and how far above \(L\) the middle function can be.
Applying the Theorem
Worked Example: A Nonnegative Quotient Near Zero
Evaluate \(\lim_{x\to 0}\frac{x^2}{1+x^2}\).
For every real \(x\), \(1+x^2\ge1>0\), so the denominator is positive. Also \(x^2\ge0\), which gives
Since \(1+x^2\ge1\), its reciprocal satisfies \(\frac{1}{1+x^2}\le1\). Multiplying by \(x^2\ge0\) preserves the inequality, so
As \(x\to0\), the lower bound \(0\) approaches \(0\), and the upper bound \(x^2\) also approaches \(0\). Both bounds hold for every \(x\), in particular throughout every punctured neighborhood of \(0\). The Squeeze Theorem therefore gives
Check: The denominator tends to \(1\), while the numerator tends to \(0\); direct substitution also gives \(0/1=0\), consistent with the squeezed limit.
In that example, the function was already straightforward to evaluate. The theorem is especially helpful when a function has no simple limit of its own in a factor, but the full expression is still trapped by bounds that converge together. The next example makes the bounding step do the essential work.
Worked Example: A Bounded Factor Multiplied by a Vanishing Factor
Define \(q(x)=2\) when \(x\) is rational and \(q(x)=-1\) when \(x\) is irrational, and let \(f(x)=x^2q(x)\). Find \(\lim_{x\to0}f(x)\).
The factor \(q(x)\) does not settle toward a single value as \(x\) approaches \(0\): it takes the values \(2\) and \(-1\) arbitrarily close to \(0\). But for every real \(x\), \(-1\le q(x)\le2\). Because \(x^2\ge0\), multiplying this inequality by \(x^2\) gives
Both bounds approach \(0\), since \(\lim_{x\to0}(-x^2)=0\) and \(\lim_{x\to0}2x^2=0\). The inequalities hold for every \(x\), including throughout a punctured neighborhood of \(0\). Therefore,
Check: At rational inputs the expression is \(2x^2\), which tends to \(0\). At irrational inputs it is \(-x^2\), which also tends to \(0\). These are exactly the two bounds, confirming that the expression is trapped near \(0\) for either kind of input.
Worked Example: Squeezing Toward a Nonzero Value
For \(x\ne2\), define \(F(x)=4+\frac{(x-2)^2(x+1)}{x^2+1}\). Evaluate \(\lim_{x\to2}F(x)\) using bounds.
Restrict attention to \(0<|x-2|<1\), so \(1<x<3\). Then \(x+1>0\), \(x+1<4\), and \(x^2+1\ge1\). Consequently,
Since \((x-2)^2\ge0\), multiplication gives
Adding \(4\) to each part yields the needed two-sided bounds:
As \(x\to2\), the lower bound approaches \(4\), and the upper bound approaches \(4+4(0)^2=4\). The bounds hold whenever \(0<|x-2|<1\), so the theorem applies and gives
Check: The term added to \(4\) is nonnegative and at most \(4(x-2)^2\), which tends to \(0\). Thus that term vanishes in the limit, agreeing with the result.
Conditions to Check Before Squeezing
Before applying the theorem, verify each part of its hypothesis rather than relying on the visual appearance of an inequality.
- Both bounds: Find functions \(g\) and \(h\) that constrain \(f\) from below and above.
- Nearby inputs: Confirm that \(g(x)\le f(x)\le h(x)\) holds throughout a punctured neighborhood of the point.
- Common limit: Evaluate the limits of both bounds and verify that they are the same finite number.
- Two-sided approach: The inequalities and bound limits must apply as \(x\) approaches from either side, not just from one side.
The inequalities do not need to hold at \(x=a\). The function may even be undefined there, as in the third example. What matters is that the conditions hold for all sufficiently close \(x\ne a\).
A Common Mistake: Bounds That Do Not Hold Nearby
A limit cannot be concluded from inequalities that are false near the point. For instance, let \(f(x)=1\) for \(x\ne0\). It would be tempting to write
and claim that both bounds approach \(0\), so \(f(x)\to0\). But the right-hand inequality is false whenever \(0<|x|<1\): then \(x^2<1\). The proposed squeeze does not apply. In fact, the valid bounds \(1\le f(x)\le1\) hold for every \(x\ne0\), and both bounds approach \(1\), so the correct limit is \(1\).
A related error is checking the inequality only at a few sample inputs. A squeeze requires a neighborhood-wide guarantee. Testing examples may help you look for a bound, but it cannot replace proving that the bound holds for every nearby input.
Check Your Understanding
For each question, focus on whether all the Squeeze Theorem hypotheses are present.
- State the Squeeze Theorem for a two-sided limit, including where the inequalities must hold.
- If \(0\le f(x)\le 3x^2\) for every \(x\ne0\), what can you conclude about \(\lim_{x\to0}f(x)\)? Why?
- Why is it not necessary for the squeeze inequalities to hold at \(x=a\)?
- If the lower bound approaches \(2\) but the upper bound approaches \(5\), does the Squeeze Theorem determine the limit of the middle function? Explain.
- For \(F(x)=4+\frac{(x-2)^2(x+1)}{x^2+1}\), what neighborhood condition in the worked example made the bound on \(\frac{x+1}{x^2+1}\) valid?
