Why the Order of the Bounds Matters
A squeeze argument is not justified merely because two expressions approach the same number as \(x\) approaches \(a\). At every relevant input near \(a\), the proposed lower bound must actually be less than or equal to the function, and the function must be less than or equal to the proposed upper bound.
This pointwise order can fail when an expression changes sign. It can also be meaningless on part of a two-sided neighborhood if the function or a proposed bound is not defined there. Before using the Squeeze Theorem, check the inequalities where the limit is being taken.
The word throughout is essential. The inequalities must hold for every \(x\) close enough to \(a\) in the relevant domain, not just for selected inputs or on one side when a two-sided limit is claimed.
A Reliable Order Check
To check whether \(g(x)\leq f(x)\leq h(x)\), test the two inequalities separately. Equivalently, verify that both differences \(f(x)-g(x)\) and \(h(x)-f(x)\) are nonnegative throughout the neighborhood. If either difference can change sign, the proposed squeeze may fail.
Pay particular attention to operations that can affect order:
- Multiplying an inequality by a positive quantity preserves its direction; multiplying by a negative quantity reverses it. If the multiplier changes sign, handle the different sign cases separately.
- Dividing by a quantity requires checking that it is nonzero and determining its sign. A denominator that is positive on one side of \(a\) might be negative on the other.
- An absolute value is always nonnegative. In particular, \(-|u(x)|\leq u(x)\leq |u(x)|\), regardless of the sign of \(u(x)\).
- Every expression in the inequalities must be defined on the neighborhood or one-sided neighborhood where the squeeze is being used.
A useful habit is to write the neighborhood explicitly, such as \(0<|x-a|<\delta\) for a two-sided limit or \(a<x<a+\delta\) for a right-hand limit. Then check the order under exactly those conditions.
Why the Squeeze Theorem Works
The order condition connects the limits of the bounds to the values of the function between them. Here is the short justification for the two-sided version. Suppose both bounds tend to \(L\), and choose any \(\varepsilon>0\). By their limits, sufficiently close to \(a\), both bounds lie between \(L-\varepsilon\) and \(L+\varepsilon\). The pointwise inequalities then force \(f(x)\) into that same interval.
Thus \(|f(x)-L|<\varepsilon\) for every sufficiently close \(x\ne a\). Since this holds for every \(\varepsilon>0\), \(f(x)\) tends to \(L\). If the inequalities fail, this chain of reasoning cannot be made: the bounds do not necessarily constrain \(f\).
Worked Example: A Factor That Changes Sign
Worked Example: A Factor That Changes Sign
Use a valid squeeze to find \(\lim_{x\to 0}x\sin(1/x)\), and check why a tempting tighter pair of bounds is invalid.
For every real \(u\), \(-1\leq\sin u\leq 1\), so \(|\sin u|\leq 1\). It follows that \(|x\sin(1/x)|\leq |x|\). The absolute-value inequality gives both bounds:
These bounds are ordered because \(-|x|\leq |x|\). Both tend to zero as \(x\to 0\), so the Squeeze Theorem gives \(\lim_{x\to 0}x\sin(1/x)=0\).
Check the inequalities by cases. If \(x>0\), multiplying \(-1\leq\sin(1/x)\leq 1\) by \(x\) gives \(-x\leq x\sin(1/x)\leq x\). If \(x<0\), multiplying by \(x\) reverses both inequality signs, giving \(x\leq x\sin(1/x)\leq -x\). These are precisely the bounds using \(|x|\) on each side.
A common error is to claim \(-x\leq x\sin(1/x)\leq x\) for both positive and negative \(x\). For \(x<0\), the proposed lower bound \(-x\) is greater than the proposed upper bound \(x\), so the bounds are not even in lower-to-upper order. Another tempting claim is \(-x^2\leq x\sin(1/x)\leq x^2\). It fails, for example, at \(x=1/(\pi/2+2\pi n)\) for every nonnegative integer \(n\) large enough that \(0<x<1\): here \(\sin(1/x)=1\), so \(x\sin(1/x)=x>x^2\). The corrected bounds \(-|x|\) and \(|x|\) are valid throughout the punctured neighborhood and are sufficient to establish the limit.
Check: The middle expression lies between the stated bounds for both signs of \(x\), and each bound has limit zero. Therefore the squeeze conclusion applies.
One-Sided Domains and One-Sided Squeezes
A two-sided limit concerns inputs on both sides of the limit point. But some functions are defined near a point only on one side. In that case, a one-sided limit can still be studied: the function and inequalities need to hold on the relevant side. Their failure to be defined on the other side does not invalidate a one-sided squeeze, but it does prevent using that squeeze to claim a two-sided limit.
Worked Example: A Function Defined Only to the Right
For \(x>0\), let \(f(x)=\sqrt{x}\sin(1/x)\). Find its right-hand limit at zero and explain why this argument does not establish a two-sided limit.
The function is real-valued for \(x>0\). Since \(-1\leq\sin(1/x)\leq 1\), and \(\sqrt{x}>0\) on this domain, multiplication preserves the inequality direction:
The lower and upper bounds both tend to zero as \(x\to 0^+\). Therefore the one-sided Squeeze Theorem gives \(\lim_{x\to 0^+}\sqrt{x}\sin(1/x)=0\).
Check the domain and order: for every \(x>0\), \(\sqrt{x}\) exists as a real number and is positive, so the multiplication of the sine inequality is valid. The function and both bounds are defined, and ordered, throughout the right-hand punctured neighborhood.
This does not prove a two-sided limit. For \(x<0\), \(\sqrt{x}\) is not real-valued, so \(f\) is not defined on a left-hand neighborhood of zero. The missing left side is a domain issue, not a failure of the right-hand squeeze.
Division and Bounds That Change Sign
When dividing an inequality, checking the divisor’s sign is just as important as checking the original order. A useful way to avoid a sign split is to divide by a positive quantity such as an absolute value, when that quantity is nonzero. The next example uses this idea to bound an expression whose sign changes at the limit point.
Worked Example: Dividing by an Absolute Value
Find \(\lim_{x\to 0}\frac{1-\cos x}{x}\), checking the order on both sides of zero.
Use \(1-\cos x=2\sin^2(x/2)\) and the local sine bound \(|\sin u|\leq |u|\). The identity shows that the numerator is nonnegative. The sine bound gives
For \(x\ne 0\), divide by \(|x|\), which is positive. This gives \(0\leq (1-\cos x)/|x|\leq |x|/2\). Since \(\left|(1-\cos x)/x\right|=(1-\cos x)/|x|\), the absolute-value bound is
The bounds are ordered, and both tend to zero as \(x\to 0\). The Squeeze Theorem therefore gives \(\lim_{x\to 0}(1-\cos x)/x=0\).
A common mistake is to divide \(0\leq 1-\cos x\leq x^2/2\) by \(x\) and keep the inequality directions unchanged on both sides of zero. That is only valid when \(x>0\). When \(x<0\), division by \(x\) reverses the directions, and the quotient \((1-\cos x)/x\) is nonpositive. For instance, at \(x=-1/2\), the numerator \(1-\cos(-1/2)\) is positive while the denominator is negative, so the quotient is negative; a claimed lower bound of zero is false. Dividing by \(|x|\) avoids that error, and the final symmetric bounds handle both signs.
Check: The divisor \(|x|\) is positive for every \(x\ne 0\); the final lower bound is at most the upper bound; and both bounds approach zero. Each part of the squeeze is valid on the full punctured neighborhood.
A Practical Audit Before Concluding
Before applying the Squeeze Theorem, pause and verify the proposed bounds rather than relying on a familiar-looking pattern. In particular, make sure you have not used a comparison that is true only for positive inputs when the limit is two-sided.
Decide whether the claim is two-sided, a right-hand limit, or a left-hand limit.
Confirm that the function and both bounds are defined on the relevant punctured neighborhood.
Check \(g(x)\leq f(x)\) and \(f(x)\leq h(x)\) throughout that neighborhood, accounting for sign changes.
Only after the domain and order checks, verify that the lower and upper bounds tend to the same value.
The order check is what makes the squeeze an argument rather than a guess. If a bound fails, revise it—for example, by using an absolute value or restricting the claim to a valid one-sided domain—before drawing a conclusion.
Check Your Understanding
For each question, identify the domain and the relevant side of the limit before deciding whether the bounds are valid.
- Why does \(-x\leq x\sin(1/x)\leq x\) fail as a two-sided bound near zero?
- What must be checked before multiplying an inequality by a factor that may be negative?
- Why can a valid right-hand squeeze for a function undefined to the left of \(a\) not establish its two-sided limit at \(a\)?
- When dividing an inequality by \(x\), what changes between the cases \(x>0\) and \(x<0\)?
- For a proposed squeeze \(g(x)\leq f(x)\leq h(x)\), what two pointwise inequalities should be checked before examining the limits of \(g\) and \(h\)?
