When Bounds Determine a Limit
Sometimes a function is difficult to evaluate directly near a point, but it is easier to compare it with two other functions. If one function stays below it, another stays above it, and both bounding functions approach the same number, then the middle function has no room to approach anything else.
This principle is called the Squeeze Theorem. Its force comes from a simple order fact: if a number lies between two numbers that are both close to \(L\), then it too is close to \(L\). The theorem applies that fact to function values near an input \(a\).
Statement of the Squeeze Theorem
The comparison is required near \(a\), not necessarily at \(a\). In particular, the functions involved need not be defined at \(a\). A punctured neighborhood of \(a\) means an interval around \(a\) with the point \(a\) itself removed.
The inequality must hold throughout a punctured neighborhood: an inequality at only a few selected inputs is not enough. Both outer functions must also have limits, and those limits must be the same finite number. No assumption is made about the value \(f(a)\), even if that value exists.
Why the Theorem Works
Here is a proof directly from the definition of a limit. For a finite limit \(f(x)\to L\) as \(x\to a\), the requirement is that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\varepsilon\).
Let \(\varepsilon>0\) be arbitrary. Since \(g(x)\to L\), there is a positive \(\delta_g\) such that \(0<|x-a|<\delta_g\) implies \(L-\varepsilon<g(x)<L+\varepsilon\).
Since \(h(x)\to L\), there is a positive \(\delta_h\) such that \(0<|x-a|<\delta_h\) implies \(L-\varepsilon<h(x)<L+\varepsilon\).
Choose \(\delta\) no larger than \(\delta_g\), \(\delta_h\), and the radius of the neighborhood where \(g(x)\leq f(x)\leq h(x)\). Then all these statements hold whenever \(0<|x-a|<\delta\).
For those \(x\), the inequalities give \(L-\varepsilon<g(x)\leq f(x)\leq h(x)<L+\varepsilon\). Therefore \(L-\varepsilon<f(x)<L+\varepsilon\), so \(|f(x)-L|<\varepsilon\).
Because this works for every \(\varepsilon>0\), the definition of a limit gives \(\lim_{x\to a}f(x)=L\). The proof uses order to transfer closeness from the two bounding functions to the function between them.
Applying the Theorem
A useful application begins by proving the inequalities, not by assuming them from a graph or a few function values. Once the pointwise bounds are established, the outer limits can be calculated. If both equal \(L\), the theorem supplies the limit of the middle function.
Worked Example: A Rational Function Bounded by a Square
Find \(\lim_{x\to 0}\frac{x^2}{1+x^2}\). For every real \(x\), \(x^2\geq 0\), so \(1+x^2\geq 1\). The denominator is positive. Thus
The lower function has limit \(\lim_{x\to 0}0=0\). The upper function has limit \(\lim_{x\to 0}x^2=0\). Both bounds approach the same number, and they enclose the given function near \(0\). By the Squeeze Theorem,
Check: the proved inequalities hold for every real \(x\), including every \(x\neq 0\) close to \(0\), and the two outer limits both equal \(0\). These are exactly the theorem's requirements.
Worked Example: A Squeezed Limit That Is Not Zero
Find \(\lim_{x\to 2}\left(3+\frac{(x-2)^2}{1+(x-2)^2}\right)\). Since \((x-2)^2\geq 0\), its denominator satisfies \(1+(x-2)^2\geq 1\). Therefore
Adding \(3\) to all three parts preserves the order:
As \(x\to 2\), the lower function has limit \(3\), and the upper function has limit \(3+0^2=3\). The middle function is therefore squeezed to \(3\):
Check: substituting \(x=2\) into the expression gives \(3+0/(1+0)=3\), consistent with the limit. The squeeze argument establishes the limit from bounds in a punctured neighborhood; it does not require the value at \(2\).
Bounds Can Handle Functions Defined by Cases
The function being squeezed does not need to have one formula throughout its domain. The theorem needs the order bounds to hold near the point, not a particular algebraic form for the middle function.
Worked Example: A Limit for a Two-Case Function
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^2\) when \(x\) is rational and \(f(x)=2x^2\) when \(x\) is irrational. Find \(\lim_{x\to 0}f(x)\).
For rational \(x\), \(f(x)=x^2\), and \(x^2\leq 2x^2\) because \(x^2\geq 0\). For irrational \(x\), \(f(x)=2x^2\), so it also satisfies \(x^2\leq f(x)\leq 2x^2\). Thus the bounds hold for every real \(x\):
The lower limit is \(\lim_{x\to 0}x^2=0\), and the upper limit is \(\lim_{x\to 0}2x^2=2\cdot 0=0\). The Squeeze Theorem gives
Check: for either case, the value of \(f(x)\) lies between \(x^2\) and \(2x^2\), and both bounds approach \(0\). The conclusion is valid even though the formula used for \(f\) depends on whether \(x\) is rational or irrational.
What the Hypotheses Do—and Do Not—Say
The theorem is a sufficient condition for a limit: if the inequalities and matching outer limits hold, then the middle limit follows. It does not say that every useful pair of bounds must have matching limits, nor that unequal limiting bounds prove the middle function has no limit. It says only that this particular squeeze argument cannot establish the conclusion when its hypotheses are missing.
For example, define \(q(x)=1\) for \(x>0\) and \(q(x)=-1\) for \(x<0\), for \(x\neq 0\). Then \(-1\leq q(x)\leq 1\) near \(0\), but the outer constants have different limits, \(-1\) and \(1\). The bounds do not squeeze \(q\) to one number: \(q\) equals \(1\) on the positive side and \(-1\) on the negative side, so it has no two-sided limit at \(0\). The matching-limit condition is essential to this theorem.
A second common error is reversing an inequality after multiplying or dividing by a negative quantity. The order relation must be preserved correctly at every step. In the examples above, the denominators are positive and the quantities being compared are nonnegative, so the displayed bounds have the stated direction.
Why Order Bounds Matter
Limits describe the behavior of function values arbitrarily close to an input. Pointwise inequalities add information about how those values can vary: the middle value cannot pass below its lower bound or above its upper bound. If both restrictions close in on the same number, then every permitted middle value must close in on that number as well.
This perspective is useful when direct substitution gives no answer or when the middle function changes form too irregularly for an algebraic simplification. The key work is to find valid bounds whose own limits are manageable. The theorem then turns those comparisons into a rigorous limit conclusion.
Check Your Understanding
For each question, identify the relevant bounds and check whether the theorem's hypotheses are satisfied.
- If \(g(x)\leq f(x)\leq h(x)\) near \(a\), and both outer limits equal \(L\), what conclusion follows about \(f\)?
- Why does the proof choose a \(\delta\) that works for both outer limits and the inequality neighborhood?
- For the function \(\frac{(x-1)^2}{1+(x-1)^2}\), find simple lower and upper bounds that have matching limits as \(x\to 1\).
- Must \(f(a)\) be defined for the Squeeze Theorem to establish \(\lim_{x\to a}f(x)\)?
- Why do the bounds \(-1\leq q(x)\leq 1\) alone fail to establish a limit for the two-case function \(q\) described above at \(0\)?
