What Is The Squeeze Theorem?
The Squeeze Theorem (also called the Sandwich Theorem) is a rule for finding the limit of a function that is hard to evaluate directly, by trapping it between two functions whose limits are known and equal. Here is the precise statement.
Statement. Suppose $g(x) \le f(x) \le h(x)$ for every $x$ near $a$ (except possibly at $a$ itself). If $$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,$$ then $f$ has a limit at $a$ and $$\lim_{x \to a} f(x) = L.$$
Three parts of that statement carry all the weight, and each one is a place learners slip:
The inequality $g(x) \le f(x) \le h(x)$ only has to hold near $a$, not everywhere, and not necessarily at $a$ itself.
The two outer functions must approach the same number $L$. If they head to different values, the theorem says nothing.
The middle function $f$ is the one you could not evaluate on its own. The outer functions $g$ and $h$ are chosen precisely because they are easy.
The name is the picture: $f$ is the filling, $g$ and $h$ are the two slices of bread. Pinch the bread together at height $L$ and the filling has nowhere to go but $L$.
How Do You Use The Squeeze Theorem On $x^2 \sin\frac{1}{x}$?
The first showcase limit is one that resists every ordinary method. As $x \to 0$, the factor $\sin\frac{1}{x}$ oscillates faster and faster between $-1$ and $1$ and never settles, so you cannot substitute, factor, or cancel. The Squeeze Theorem handles it in three steps.
Step 1: Bound the oscillating factor.
For every real input, sine stays between $-1$ and $1$:
$$-1 \le \sin\frac{1}{x} \le 1$$
Step 2: Multiply through by $x^2$.
Because $x^2 \ge 0$, multiplying does not flip the inequality:
$$-x^2 \le x^2 \sin\frac{1}{x} \le x^2$$
Now the middle expression is exactly the function we want, trapped between $g(x) = -x^2$ and $h(x) = x^2$.
Step 3: Take the limits of the two bounds.
$$\lim_{x \to 0} (-x^2) = 0 \qquad \text{and} \qquad \lim_{x \to 0} x^2 = 0$$
Both outer functions approach the same value, $L = 0$. By the Squeeze Theorem:
$$\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$$
Final answer: $\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$.
The geometric reading is worth pausing on. The graph of $x^2 \sin\frac{1}{x}$ wobbles wildly, but it lives entirely inside the two parabolas $y = x^2$ and $y = -x^2$, which pinch shut at the origin. No matter how violently the curve oscillates, the closing gap between the parabolas drags it to zero.
How Do You Prove $\lim_{x \to 0}\frac{\sin x}{x} = 1$?
The second showcase limit is the one every calculus course leans on, because it is the seed of the derivative of sine and, through it, all differentiation of trigonometric functions. Substituting $x = 0$ gives the indeterminate form $\frac{0}{0}$, so we bound the ratio geometrically.
Picture a unit circle and a small positive angle $x$ (measured in radians) at the centre. Compare three regions built on that angle:
The inner triangle, with area $\tfrac{1}{2}\sin x$.
The circular sector, with area $\tfrac{1}{2}x$.
The outer triangle, with area $\tfrac{1}{2}\tan x$.
Because each region sits inside the next, their areas are ordered:
$$\tfrac{1}{2}\sin x \le \tfrac{1}{2}x \le \tfrac{1}{2}\tan x$$
Multiply through by $2$ and use $\tan x = \frac{\sin x}{\cos x}$:
$$\sin x \le x \le \frac{\sin x}{\cos x}$$
Divide every part by $\sin x$, which is positive for $0 < x < \tfrac{\pi}{2}$:
$$1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}$$
Take reciprocals, which reverses the inequalities:
$$\cos x \le \frac{\sin x}{x} \le 1$$
Now the ratio $\frac{\sin x}{x}$ is squeezed between $g(x) = \cos x$ and $h(x) = 1$. As $x \to 0$:
$$\lim_{x \to 0} \cos x = 1 \qquad \text{and} \qquad \lim_{x \to 0} 1 = 1$$
Both bounds close on $L = 1$. Because $\frac{\sin x}{x}$ is an even function (top and bottom are both odd, so the ratio is even), the same bound holds for small negative $x$, giving the two-sided result:
$$\lim_{x \to 0}\frac{\sin x}{x} = 1$$
Final answer: $\lim_{x \to 0}\frac{\sin x}{x} = 1$.
Geometrically, this says that for a tiny angle the arc length $x$, the chord-related quantity $\sin x$, and the tangent segment $\tan x$ all become indistinguishable, which is exactly why a small slice of a curve looks straight.
What Is $\lim_{x \to 0}\frac{1 - \cos x}{x}$?
This limit follows directly from the last one and shows up constantly alongside it. Substituting $x = 0$ again gives $\frac{0}{0}$, so multiply the top and bottom by the conjugate $1 + \cos x$:
$$\frac{1 - \cos x}{x} = \frac{(1 - \cos x)(1 + \cos x)}{x,(1 + \cos x)} = \frac{1 - \cos^2 x}{x,(1 + \cos x)}$$
Use the identity $1 - \cos^2 x = \sin^2 x$ and split the fraction:
$$\frac{\sin^2 x}{x,(1 + \cos x)} = \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x}$$
Now take the limit of each piece as $x \to 0$. The first piece is the result we just proved, $\frac{\sin x}{x} \to 1$. The second piece tends to $\frac{0}{1 + 1} = 0$:
$$\lim_{x \to 0}\frac{1 - \cos x}{x} = 1 \cdot \frac{0}{2} = 0$$
Final answer: $\lim_{x \to 0}\frac{1 - \cos x}{x} = 0$.
What Are Some Worked Squeeze Theorem Examples?
Example 1: A bounded factor times a vanishing one.
Evaluate $\lim_{x \to 0} x^2 \cos\frac{1}{x}$.
Cosine is bounded the same way sine is:
$$-1 \le \cos\frac{1}{x} \le 1$$
Multiply by $x^2 \ge 0$:
$$-x^2 \le x^2 \cos\frac{1}{x} \le x^2$$
Both outer limits are $0$, so the middle is squeezed to $0$.
Final answer: $\lim_{x \to 0} x^2 \cos\frac{1}{x} = 0$.
Example 2: A three-function bound at a non-zero point.
Suppose a function $f$ satisfies $2x \le f(x) \le x^2 + 1$ for all $x$ near $1$. Find $\lim_{x \to 1} f(x)$.
Evaluate each bound at $a = 1$:
$$\lim_{x \to 1} 2x = 2 \qquad \text{and} \qquad \lim_{x \to 1} (x^2 + 1) = 2$$
Both bounds meet at $L = 2$, and $f$ is trapped between them, so:
Final answer: $\lim_{x \to 1} f(x) = 2$.
Example 3: Squeezing a limit at infinity.
Evaluate $\lim_{x \to \infty}\frac{\sin x}{x}$.
The numerator never leaves the band $-1 \le \sin x \le 1$, so for $x > 0$:
$$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$$
As $x \to \infty$, both $-\frac{1}{x}$ and $\frac{1}{x}$ tend to $0$. The Squeeze Theorem gives:
Final answer: $\lim_{x \to \infty}\frac{\sin x}{x} = 0$.
Note that this is a different limit from $\lim_{x \to 0}\frac{\sin x}{x} = 1$. Same expression, different approach point, different answer, which is a useful reminder that a limit is always tied to where $x$ is heading.
Which Bounds Should You Reach For? (A Quick Reference)
Most Squeeze Theorem problems fall into a few recognisable shapes. This table pairs the tell-tale form with the bound that usually cracks it.
Table: Common Squeeze Theorem setups and the bounds that solve them.
If the function looks like… | Use the bound… | Because it tends to… |
|---|---|---|
$x^2 \sin\frac{1}{x}$, $x^2 \cos\frac{1}{x}$ | $-x^2 \le \text{(it)} \le x^2$ | $0$ |
$x \sin\frac{1}{x}$ (near $0$) | $-\lvert x\rvert \le \text{(it)} \le \lvert x\rvert$ | $0$ |
$\dfrac{\sin x}{x}$ (near $0$) | $\cos x \le \dfrac{\sin x}{x} \le 1$ | $1$ |
$\dfrac{\sin x}{x}$ (as $x \to \infty$) | $-\dfrac{1}{x} \le \dfrac{\sin x}{x} \le \dfrac{1}{x}$ | $0$ |
any bounded factor $\times$ a factor $\to 0$ | $-\lvert\text{factor}\rvert \le \text{(it)} \le \lvert\text{factor}\rvert$ | $0$ |
The pattern in the last row is the master key: whenever a term stays inside a fixed band and multiplies something that vanishes, the product is squeezed to zero.
Why Does The Squeeze Theorem Work?
The theorem is one of the rare results in calculus that is as obvious in a picture as it is airtight in a proof. Here is the intuition, kept to its parts:
Near $a$, the graph of $f$ is boxed in from below by $g$ and from above by $h$. It cannot cross either boundary.
As $x \to a$, the top and bottom of that box close on the single height $L$. The vertical gap the graph is allowed to occupy shrinks to nothing.
A value pinned inside a gap that shrinks to a point has only one place left to be. So $f(x)$ is forced to $L$.
The rigorous version replaces "closes on" with the epsilon-delta definition of a limit. Because $g \to L$ and $h \to L$, for any tolerance we can force both $g(x)$ and $h(x)$ within that tolerance of $L$; the inequality $g(x) \le f(x) \le h(x)$ then drags $f(x)$ into the same tolerance. That is the whole proof, and it is why the result is not just plausible but certain.
One thing the intuition also makes clear: the theorem needs both bounds to arrive at the same $L$. If the box never closes, because the top and bottom head to different heights, $f$ is free to wander inside the gap and may have no limit at all.
Who Discovered The Squeeze Theorem?
The idea is far older than the calculus notation we write it in. It grew out of the ancient Greek method of measuring curved shapes by trapping them between straight-sided ones.
The modern, symbolic version came with the nineteenth-century drive to make calculus rigorous. Carl Friedrich Gauss (1777–1855, Germany) stated the bounding argument in modern terms, and Augustin-Louis Cauchy (1789–1857, France) supplied the epsilon-based definition of a limit that turns the geometric picture into a formal proof.
Where Is The Squeeze Theorem Used In The Real World?
Its reach goes well past the exam question, because "trap the unknown between two knowns" is a strategy engineers and scientists use constantly.
Signal processing: a noisy signal that stays inside a decaying envelope (like $x^2 \sin\frac{1}{x}$ inside $\pm x^2$) is guaranteed to settle, which is how engineers prove a filtered signal dies out.
Numerical error bounds: when an exact value cannot be computed, it is squeezed between an upper and lower estimate that converge, giving a guaranteed accuracy, the same logic Archimedes used for $\pi$.
Physics of oscillations: a damped vibration bounded by a shrinking amplitude curve is shown to stop using exactly this bounding argument.
Machine learning: convergence proofs for training algorithms routinely trap an error term between two sequences that both tend to zero.
Computer graphics and animation: motion pinned between two easing curves that meet is smoothed to a predictable endpoint.
One habit, bounding a hard thing between two easy ones, runs from an ancient measurement of the circle to the algorithms training today's models.
What Are The Most Common Mistakes With The Squeeze Theorem?
These three errors account for most lost marks, and each traces back to dropping one of the theorem's conditions.
Using bounds that do not share the same limit.
Where it slips in:
A student finds some lower bound and some upper bound, checks that the function sits between them, and declares victory, without confirming both bounds head to the same number.
Don't do this:
Do not conclude anything when $\lim g = 0$ but $\lim h = 1$. Trapping a function between $0$ and $1$ only proves the limit is somewhere in that range, not that it exists.
The correct way:
Compute $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ separately and confirm they are equal before applying the theorem. Only their common value $L$ can be the answer.
Forgetting the inequality only needs to hold near $a$.
Where it slips in:
A student abandons a valid bound because the inequality fails at one far-away point, or at $a$ itself, and assumes the theorem no longer applies.
Don't do this:
Do not require $g(x) \le f(x) \le h(x)$ for every $x$. The theorem only asks that it hold on some interval around $a$, and the value exactly at $a$ never matters for a limit.
The correct way:
Check the inequality only on a small neighbourhood of $a$ (excluding $a$ if needed). A bound that holds close to the point is enough, even if it breaks down elsewhere.
Mis-bounding an oscillating factor when a sign is negative.
Where it slips in:
Starting from $-1 \le \sin\frac{1}{x} \le 1$, a student multiplies by a quantity that can be negative and keeps the inequality pointing the same way.
Don't do this:
Do not multiply an inequality by a negative number without reversing it. Multiplying $-1 \le \sin\frac{1}{x} \le 1$ by $x^2$ is safe because $x^2 \ge 0$; multiplying by a factor that changes sign is not.
The correct way:
Bound with absolute values to stay safe: $\left\lvert x \sin\frac{1}{x}\right\rvert \le \lvert x\rvert$, which gives $-\lvert x\rvert \le x\sin\frac{1}{x} \le \lvert x\rvert$. Both outer bounds tend to $0$ regardless of the sign of $x$.
Practice Problems On The Squeeze Theorem
Work each limit, then check against the answer.
$\lim_{x \to 0} x^2 \sin\frac{5}{x}$.
(Answer: $0$, since $-x^2 \le x^2\sin\frac{5}{x} \le x^2$.)$\lim_{x \to 0} x \cos\frac{1}{x}$.
(Answer: $0$, since $-\lvert x\rvert \le x\cos\frac{1}{x} \le \lvert x\rvert$.)$\lim_{x \to 0}\frac{\sin 3x}{x}$.
(Answer: $3$, write it as $3\cdot\frac{\sin 3x}{3x}$ and use $\frac{\sin u}{u} \to 1$.)A function obeys $4 - x^2 \le f(x) \le 4 + x^2$ near $x = 0$. Find $\lim_{x \to 0} f(x)$.
(Answer: $4$, both bounds tend to $4$.)$\lim_{x \to \infty}\frac{\cos x}{x}$.
(Answer: $0$, since $-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}$.)$\lim_{x \to 0}\frac{1 - \cos x}{x^2}$.
(Answer: $\frac{1}{2}$, rewrite as $\frac{\sin^2 x}{x^2(1+\cos x)} = \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1+\cos x} \to 1\cdot\frac{1}{2}$.)
Where Should You Go Next After The Squeeze Theorem?
The Squeeze Theorem is a gateway into the limit-based heart of calculus, and a few natural doors open from here.
Calculus. Step back to the big picture of limits, derivatives, and integrals, and see where the squeeze fits in the whole subject.
Differentiation of trigonometric functions. The limit $\frac{\sin x}{x} \to 1$ is exactly what makes the derivative of $\sin x$ come out to $\cos x$.
Trigonometric functions. Shore up the sine and cosine facts the two showcase limits depend on.
If your child is building these foundations, a live Bhanzu trainer teaches limits starting from the "why", the geometric trapping the Squeeze Theorem is built on, in Bhanzu's online math programme.
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