Squeeze theorem (sandwich theorem): Worked examples

Before we start, one reminder in plain words. The symbol $lim_{x \to a} f (x) = L$ is read "as $x$ gets closer and closer to the number $a$ (but never equals it), the value $f (x)$ gets closer and closer to $L$ ." The arrow $\to$ means "approaches." We never arrive at $a$ ; we sneak up on it. Everything below rests on that one idea, powered underneath by the Limit definition (epsilon-delta).

The scenario matrix

Every squeeze problem falls into one of these cells. Each row is a class of difficulty; the last column names the worked example that covers it.

Cell	What makes it tricky	Covered by
A. Positive multiplier	Bounded wiggle $\times$ a factor that is always $\geq 0$	Example 1
B. Sign-changing multiplier	Multiplier can be negative near $a$ — inequality may flip	Example 2
C. Both one-sided limits	$x \to a^{+}$ and $x \to a^{-}$ behave differently; must check both sides	Example 3
D. Sequence version	$n \to \infty$ instead of $x \to a$	Example 4
E. Degenerate / walls-don't-meet	The bounds tend to different values → theorem gives nothing	Example 5
F. Limit at infinity (function)	$x \to \infty$ , bounds shrink to $0$	Example 6
G. Real-world word problem	A physical damped oscillation	Example 7
H. Exam twist	Bound is not obvious; must construct it	Example 8

We now hit every cell.

Example 1 — Cell A: positive multiplier

Worked example Evaluate

$x \to 0 lim x^{4} cos \frac{1}{x ^{2}}$ . Forecast: $cos (1/ x^{2})$ oscillates insanely fast near $0$ . Guess the limit before reading — does the wild wiggle survive, or does the $x^{4}$ crush it?

Step 1. Bound the wiggle. For every real input $t$ , $- 1 \leq cos t \leq 1$ . Here $t = 1/ x^{2}$ , so $- 1 \leq cos \frac{1}{x ^{2}} \leq 1.$ Why this step? We can't evaluate $cos (1/ x^{2})$ at $x = 0$ (its input $1/ x^{2}$ blows up), but cosine is caged between $- 1$ and $1$ no matter how crazy the input. That cage is our starting wall.

Step 2. Multiply through by $x^{4}$ . Since $x^{4} \geq 0$ for every real $x$ , the inequality direction is preserved: $- x^{4} \leq x^{4} cos \frac{1}{x ^{2}} \leq x^{4} .$ Why this step? This is exactly cell A: the multiplier $x^{4}$ is a fourth power, so it can never be negative. Multiplying an inequality by a non-negative number keeps $\leq$ as $\leq$ .

Step 3. Both walls collapse: $lim_{x \to 0} (- x^{4}) = 0$ and $lim_{x \to 0} x^{4} = 0$ . Why this step? The walls share the common limit $0$ — the one condition the parent note warns you never to skip.

Conclusion. By squeeze, $x \to 0 lim x^{4} cos \frac{1}{x ^{2}} === 0 == .$

Verify: Take $x = 0.1$ : $x^{4} = 1 0^{- 4}$ , and $cos (1/0.01) = cos (100) \approx 0.862$ , giving $\approx 8.6 \times 1 0^{- 5}$ — already tiny and pinned inside $[- 1 0^{- 4}, 1 0^{- 4}]$ . The envelope crushes the wiggle. ✓

This is the Bounded times vanishing pattern: bounded $\times$ (something $\to 0$ ) $= 0$ .

Example 2 — Cell B: the multiplier changes sign

Worked example Evaluate

$x \to 0 lim x sin \frac{1}{x}$ . Forecast: almost identical to Example 1, but the multiplier is $x$ , not $x^{4}$ . Near $0$ , $x$ is negative on the left, positive on the right. Does that break the squeeze?

Step 1. As before, $- 1 \leq sin \frac{1}{x} \leq 1$ . Why this step? Same cage — sine lives in $[- 1, 1]$ whatever its input does.

Step 2. Here is the trap. If I naively multiply by $x$ , on the left side ( $x < 0$ ) the inequality flips (multiplying an inequality by a negative number reverses $\leq$ into $\geq$ ), so I can't write $- x \leq x sin (1/ x) \leq x$ safely. The clean fix: multiply by $∣ x ∣$ , which is always $\geq 0$ . First note $sin \frac{1}{x} \leq 1$ , so $x sin \frac{1}{x} = ∣ x ∣ sin \frac{1}{x} \leq ∣ x ∣.$ Why this step? Working with absolute values sidesteps the sign question entirely — $∣ x ∣ \geq 0$ never flips anything. This is precisely the sign-flip trap the parent note warns about (multiplying an inequality near $0$ by $x$ , which may be negative); using $∣ x ∣$ dodges it.

Step 3. Rewrite the bound without the absolute value on the outside: $- ∣ x ∣ \leq x sin \frac{1}{x} \leq ∣ x ∣.$ Why this step? $∣ A ∣ \leq B$ is the same statement as $- B \leq A \leq B$ . Now we have honest walls valid on both sides of $0$ .

Step 4. $lim_{x \to 0} (- ∣ x ∣) = 0$ and $lim_{x \to 0} ∣ x ∣ = 0$ .

Conclusion. $x \to 0 lim x sin \frac{1}{x} === 0 == .$

Verify: $x = - 0.05$ : $∣ x ∣ = 0.05$ , $sin (1/ (- 0.05)) = sin (- 20) \approx 0.913$ , product $\approx - 0.0456$ , inside $[- 0.05, 0.05]$ . ✓ The negative $x$ caused no trouble because $∣ x ∣$ absorbed the sign.

Example 3 — Cell C: one-sided limits, checked separately

Figure — Squeeze theorem (sandwich theorem)

Step 1 (right side, $x \to 0^{+}$ ). Look at the figure: the unit circle, an angle $x$ measured in radians (arc length on a radius-1 circle). Three regions nest: $area of inner triangle \leq area of sector \leq area of outer triangle .$ Their areas are $\frac{1}{2} sin x$ , $\frac{1}{2} x$ , $\frac{1}{2} tan x$ respectively. So $sin x \leq x \leq tan x$ . Why this step? We need to compare $sin x$ to $x$ itself, and geometry gives us that comparison for free — the sector is squeezed between two triangles you can see in the picture.

Step 2. Divide the chain $sin x \leq x \leq tan x$ by $sin x$ (positive for $0 < x < \frac{π}{2}$ ), then take reciprocals (which flips $\leq$ to $\geq$ ): $cos x \leq \frac{s i n x}{x} \leq 1.$ Why this step? We rearrange until the middle is exactly the quantity $\frac{s i n x}{x}$ we want. Reciprocating a chain of positives reverses it, hence the flip.

Step 3. As $x \to 0^{+}$ : $cos x \to 1$ (walls meet). Squeeze $\Rightarrow lim_{x \to 0^{+}} \frac{s i n x}{x} = 1$ .

Step 4 (left side, $x \to 0^{-}$ ). The function $\frac{s i n x}{x}$ is even: replacing $x$ by $- x$ gives $\frac{s i n ( - x )}{- x} = \frac{- s i n x}{- x} = \frac{s i n x}{x}$ , unchanged. Why this step? Evenness means the graph is mirror-symmetric about the vertical axis. Concretely, for any $x < 0$ write $x = - u$ with $u > 0$ ; then $\frac{s i n x}{x} = \frac{s i n u}{u}$ , and as $x \to 0^{-}$ we have $u \to 0^{+}$ , so the left-hand values are literally the same numbers as the right-hand ones. Hence $lim_{x \to 0^{-}} \frac{s i n x}{x} = lim_{u \to 0^{+}} \frac{s i n u}{u} = 1$ too.

Step 5 (glue the two sides). A two-sided limit exists and equals $L$ exactly when both one-sided limits exist and equal that same $L$ — this is the definition of a two-sided limit unpacked. Here both sides gave $1$ , so the loop is closed. Why this step? Steps 3 and 4 only proved the right and left limits separately; this step states the rule that lets us combine them into the single two-sided statement.

Conclusion. Both one-sided limits equal $1$ , so the two-sided $x \to 0 lim \frac{sin x}{x} === 1 == .$ (See Limits of trigonometric functions.)

Verify: $x = 0.01$ : $\frac{s i n 0.01}{0.01} \approx 0.99998$ . $x = - 0.01$ : same value by evenness $\approx 0.99998$ . Both hug $1$ . ✓

Example 4 — Cell D: the sequence version

Worked example Evaluate

$n \to \infty lim \frac{( - 1 ) ^{n} + sin n}{n}$ for positive integers $n$ . Forecast: the numerator jitters unpredictably — $(- 1)^{n}$ flips $\pm 1$ , $sin n$ wanders. Does the $n$ in the denominator win?

Step 1. Bound the numerator. $- 1 \leq (- 1)^{n} \leq 1$ and $- 1 \leq sin n \leq 1$ , so adding: $- 2 \leq (- 1)^{n} + sin n \leq 2.$ Why this step? Each piece is caged in $[- 1, 1]$ ; the sum of two caged things is caged in $[- 2, 2]$ . We now have a constant bound on the whole jittery top.

Step 2. Divide by $n > 0$ (positive, no flip): $- \frac{2}{n} \leq \frac{( - 1 ) ^{n} + s i n n}{n} \leq \frac{2}{n} .$ Why this step? Dividing an inequality by a positive number keeps its direction. This is the sequence analogue of "multiply by a positive factor."

Step 3. $lim_{n \to \infty} \frac{2}{n} = 0$ and $lim_{n \to \infty} (- \frac{2}{n}) = 0$ . Why this step? Both bounding sequences vanish — the Limits of sequences version of "walls meet."

Conclusion. Squeeze $\Rightarrow n \to \infty lim \frac{( - 1 ) ^{n} + sin n}{n} === 0 == .$

Verify: $n = 100$ : numerator $= 1 + sin 100 \approx 1 - 0.506 = 0.494$ ; divided by $100$ gives $\approx 0.0049$ , snug inside $[- 0.02, 0.02]$ . ✓ For sequences " $x \to a$ " simply becomes " $n \to \infty$ ."

Example 5 — Cell E: degenerate case, walls DON'T meet

Step 1. The bounds are honest: $- 1 \leq sin \frac{1}{x} \leq 1$ for all $x \neq = 0$ . Why this step? Same cage as always. Nothing wrong with the inequality itself.

Step 2. Check the walls' limits. $lim_{x \to 0} (- 1) = - 1$ and $lim_{x \to 0} (1) = 1$ . Why this step? The parent's golden rule: the theorem concludes only when both walls tend to the same $L$ . Here the lower wall sits at $- 1$ , the upper at $+ 1$ . They never meet.

Step 3. Because the walls stay a full distance $2$ apart, there is a wide corridor for $sin \frac{1}{x}$ to roam. Look at the figure: as $x \to 0$ the graph oscillates between $- 1$ and $+ 1$ forever, faster and faster. It never settles. Why this step? No common $L$ means squeeze gives no conclusion. And indeed the limit genuinely does not exist — this is an Oscillating functions failure.

Conclusion. Squeeze theorem: no conclusion (walls differ). The true limit: $x \to 0 lim sin \frac{1}{x}$ does not exist.

Verify: Sample $x_{n} = \frac{1}{2 π n + π /2}$ gives $sin \frac{1}{x _{n}} = sin (2 π n + \frac{π}{2}) = + 1$ ; sample $x_{n} = \frac{1}{2 π n - π /2}$ gives $- 1$ . Two subsequences approach $0$ but the function heads to $+ 1$ on one and $- 1$ on the other → no single limit. ✓ This is why you must make the walls tight.

Example 6 — Cell F: limit at infinity of a function

Worked example Evaluate

$x \to \infty lim \frac{cos x}{x ^{2} + 1}$ . Forecast: now $x$ runs off to $\infty$ rather than to a finite point. Does the squeeze idea still apply? Guess the limit.

Step 1. $- 1 \leq cos x \leq 1$ for all $x$ . Why this step? Cosine is caged. The trick is identical whether $x \to$ a finite number or $x \to \infty$ — we only bound and shrink.

Step 2. Divide by $x^{2} + 1$ , which is always positive (a square plus $1$ ), so no flip: $- \frac{1}{x ^{2} + 1} \leq \frac{c o s x}{x ^{2} + 1} \leq \frac{1}{x ^{2} + 1} .$ Why this step? $x^{2} + 1 > 0$ everywhere, so this is a safe, sign-free division — no case split needed.

Step 3. As $x \to \infty$ , $x^{2} + 1 \to \infty$ , so $\frac{1}{x ^{2} + 1} \to 0$ and $- \frac{1}{x ^{2} + 1} \to 0$ . Why this step? Both walls squeeze down to $0$ — the definition of a limit "at infinity" is the same closing-in idea.

Conclusion. $x \to \infty lim \frac{cos x}{x ^{2} + 1} === 0 == .$

Verify: $x = 10$ : $cos 10 \approx - 0.839$ , $x^{2} + 1 = 101$ , ratio $\approx - 0.0083$ , inside $[- 0.0099, 0.0099]$ . ✓

Example 7 — Cell G: real-world word problem (damped vibration)

Step 1. Cage the buzz: $- 1 \leq cos (50 t) \leq 1$ for all $t$ . Why this step? However fast the string vibrates, cosine stays in $[- 1, 1]$ . The frequency $50$ is irrelevant to the cage.

Step 2. Multiply by $e^{- t}$ . Exponentials are always positive ( $e^{- t} > 0$ for every $t$ ), so the direction is preserved: $- e^{- t} \leq e^{- t} cos (50 t) \leq e^{- t} .$ Why this step? This is cell A again in disguise — $e^{- t}$ is our non-negative multiplier. Why an exponential here and not, say, $1/ t$ ? Because physical damping (friction, air resistance) removes a fixed fraction of energy per unit time, and the function that shrinks by a constant fraction each step is precisely the exponential $e^{- t}$ .

Step 3. As $t \to \infty$ , $e^{- t} \to 0$ and $- e^{- t} \to 0$ (the walls, an envelope, collapse — see the two dashed curves hugging the wiggle in the figure). Why this step? Both walls tend to the same value $0$ .

Conclusion. $t \to \infty lim e^{- t} cos (50 t) === 0 ==$ metres — the string comes to rest at its equilibrium position.

Verify: At $t = 5$ : $e^{- 5} \approx 0.00674$ , $cos (250) \approx - 0.087$ , so $y \approx - 0.00059$ m — under a millimetre in size, trapped inside $[- 0.00674, 0.00674]$ . Units are metres throughout. ✓ The vibration hasn't stopped, but its size has shrunk to nothing.

Example 8 — Cell H: exam twist (you must construct the bound)

Worked example Evaluate

$x \to 0 lim \frac{x ^{2}}{2 + sin \frac{1}{x}}$ . Forecast: the wiggle is in the denominator now, not multiplied on top. There's no obvious "bounded $\times$ vanishing" here. Guess whether it's still $0$ .

Step 1. Control the denominator. Since $- 1 \leq sin \frac{1}{x} \leq 1$ , adding $2$ gives $1 \leq 2 + sin \frac{1}{x} \leq 3.$ Why this step? The exam trick is to notice the denominator, though wobbly, is always at least $1$ (so never zero, never negative). That saves us: we can bound the whole fraction from its denominator.

Step 2. Because the denominator lies in $[1, 3]$ and $x^{2} \geq 0$ , the fraction satisfies $\frac{x ^{2}}{3} \leq \frac{x ^{2}}{2 + s i n \frac{1}{x}} \leq \frac{x ^{2}}{1} = x^{2} .$ Why this step? A bigger denominator makes a positive fraction smaller; a smaller denominator makes it larger. Dividing $x^{2}$ by the largest denominator ( $3$ ) gives the lower wall, by the smallest ( $1$ ) gives the upper wall. Both walls are non-negative because $x^{2} \geq 0$ .

Step 3. Both walls collapse: $lim_{x \to 0} \frac{x ^{2}}{3} = 0$ and $lim_{x \to 0} x^{2} = 0$ . Why this step? The two constructed walls share the common limit $0$ — the one condition the squeeze theorem demands.

Conclusion. By squeeze, $x \to 0 lim \frac{x ^{2}}{2 + sin \frac{1}{x}} === 0 == .$

Verify: $x = 0.1$ : $sin (10) \approx - 0.544$ , denominator $\approx 1.456$ , fraction $\approx 0.01/1.456 \approx 0.00687$ , inside $[\frac{0.01}{3}, 0.01] = [0.00333, 0.01]$ . ✓

Recall Quick self-test on the matrix

Which cell has NO conclusion, and why? ::: Cell E — the two walls tend to different values ( $- 1$ and $+ 1$ ), so squeeze says nothing; the limit truly doesn't exist. In Example 2, why use $∣ x ∣$ instead of $x$ ? ::: $x$ is negative on the left of $0$ and would flip the inequality; $∣ x ∣ \geq 0$ never flips it. In Example 8, what made the denominator safe to divide by? ::: $2 + sin (1/ x) \geq 1 > 0$ always, so it's never zero or negative. Why is $e^{- t}$ (not $1/ t$ ) the right damping wall in Example 7? ::: Physical damping removes a fixed fraction of energy per unit time — that constant-fraction decay is exactly exponential. Why is proving both one-sided limits equal $1$ enough for the two-sided limit in Example 3? ::: A two-sided limit exists and equals $L$ exactly when both one-sided limits exist and equal that same $L$ .

Connections

Squeeze theorem (sandwich theorem) — the parent rule these examples exercise.
Bounded times vanishing — Examples 1, 2, 6, 7 are all this pattern.
Limits of trigonometric functions — Example 3 rebuilds $\frac{s i n x}{x} \to 1$ .
Limits of sequences — Example 4 is the $n \to \infty$ form.
Oscillating functions — Example 5 shows the failure mode.
Continuity — each limit found here can seed a continuity check.
Limit definition (epsilon-delta) — the engine under every "walls meet" step.