Exercises — Squeeze theorem (sandwich theorem)

The rule (from Squeeze theorem (sandwich theorem)): both walls must tend to the same value $L$.

• (a) ${\lim }_{x\to 0}(-x^2)=0$ and ${\lim }_{x\to 0}{x}^2=0$. Same $\Rightarrow$ YES, $f\to 0$.
• (b) The walls are the constants $2$ and $4$; they never meet. NO conclusion — $f$ could wander anywhere between $2$ and $4$.
• (c) ${\lim }_{x\to 0}(-|x|)=0$ and ${\lim }_{x\to 0}|x|=0$. Same $\Rightarrow$ YES, $f\to 0$.

"Bounded" = there are numbers $m,M$ with $m\le f(x)\le M$ everywhere.

• (a) $\sin$ of anything lies in $[-1,1]$. Bounded.
• (b) $\cos$ of anything lies in $[-1,1]$. Bounded.
• (c) $x$ grows without limit, so $x\sin x$ is not bounded (take $x=\frac{\pi }{2}+2\pi n$, value $=x\to \infty$).
• (d) Since $\sin x\in [-1,1]$, the denominator $2+\sin x\in [1,3]$, so $\frac{1}{2+\sin x}\in [\frac{1}{3},1]$. Bounded.

Step 1 — bound the wild part. $\cos$ of anything lies in $[-1,1]$: $-1\le \cos \frac{1}{x^3}\le 1.$ Why: cosine can never leave $[-1,1]$, however fast $1/x^3$ spins.

Step 2 — multiply by the tame part. Multiply through by $x^4$. Since $x^4\ge 0$ always, inequality directions are preserved (this is the Bounded times vanishing pattern): $-x^4\le x^4\cos \frac{1}{x^3}\le x^4.$

Step 3 — squeeze. ${\lim }_{x\to 0}(-x^4)=0$ and ${\lim }_{x\to 0}{x}^4=0$. Same wall-limit $\Rightarrow$ ${\lim }_{x\to 0}{x}^4\cos \frac{1}{x^3}=\boxed{0}.$

Step 1 — bound the numerator. For every real $x$, $-2\le \sin x+\cos x\le 2.$ Why: $|\sin x|\le 1$ and $|\cos x|\le 1$, so their sum is at most $2$ in size.

Step 2 — divide by $x$. For $x>0$ (we are heading to $+\infty$, so $x>0$ safely), dividing by the positive $x$ keeps directions: $-\frac{2}{x}\le \frac{\sin x+\cos x}{x}\le \frac{2}{x}.$

Step 3 — squeeze. ${\lim }_{x\to \infty }\frac{2}{x}=0$ and ${\lim }_{x\to \infty }(-\frac{2}{x})=0$. Therefore ${\lim }_{x\to \infty }\frac{\sin x+\cos x}{x}=\boxed{0}.$

This is the Limits of sequences flavour: $x\to a$ becomes $n\to \infty$.

Step 1. $(-1{)}^n$ is either $+1$ or $-1$, so $-1\le (-1{)}^n\le 1$.

Step 2 — divide by the positive $n^2+1$: $-\frac{1}{n^2+1}\le \frac{(-1{)}^n}{n^2+1}\le \frac{1}{n^2+1}.$

Step 3 — squeeze. Both walls $\pm \frac{1}{n^2+1}\to 0$. So the limit is $\boxed{0}$.

Tempting to squeeze — but check the tame factor's limit first. As $x\to 0$, $\sqrt{x^2+4}\to \sqrt{4}=2$, not $0$. The tame part does not vanish, so Bounded times vanishing does not apply.

Near $0$ the product behaves like $2\sin (1/x)$, which oscillates between about $-2$ and $2$ forever — it never settles. Any candidate walls would be roughly $-2\le f\le 2$: they do not meet, so the squeeze cannot fire.

Conclusion: the limit does not exist. (See Oscillating functions.) The lesson: squeeze only works when the walls close in; here they stay $4$ apart.

Step 1 — bound the floor. By definition of floor, for any real $t$, $t-1<\lfloor t\rfloor \le t.$ Why: the greatest integer $\le t$ sits within one unit below $t$. Put $t=1/x^2$: $\frac{1}{x^2}-1<\lfloor \frac{1}{x^2}\rfloor \le \frac{1}{x^2}.$

Step 2 — multiply by $x^2\ge 0$ (direction preserved): $x^2\left(\frac{1}{x^2}-1\right)<x^2\lfloor \frac{1}{x^2}\rfloor \le x^2\cdot \frac{1}{x^2},$ which simplifies to $1-x^2<x^2\lfloor \frac{1}{x^2}\rfloor \le 1.$

Step 3 — squeeze. Lower wall $1-x^2\to 1$, upper wall $1\to 1$. Same limit $\Rightarrow$ ${\lim }_{x\to 0}{x}^2\lfloor \frac{1}{x^2}\rfloor =\boxed{1}.$ Notice: the walls met at $1$, not $0$ — the squeeze target need not be zero.

Step 1 — restrict to a friendly window. Take $0<x<1$. Here $\ln x<0$, so $x\ln x<0$, meaning $x^x=e^{x\ln x}<e^0=1$. That is our upper wall: $x^x\le 1$.

Step 2 — lower wall. For $0<x<1$ we have $x\ln x\ge$ (something $\to 0$). Concretely, since $x\ln x\to 0$, for $x$ near $0^{+}$ the exponent lies in a shrinking interval $[-\epsilon ,0]$, so $e^{-\epsilon }\le x^x\le 1.$ As $x\to 0^{+}$ we may take $\epsilon \to 0$, giving the lower wall limit $e^0=1$.

Step 3 — squeeze. Both walls $\to 1$, so ${\lim }_{x\to 0^{+}}{x}^x=\boxed{1}$. (Cleaner route: $x^x=e^{x\ln x}$ and by Continuity of $e^t$, $\lim =e^{\lim (x\ln x)}=e^0=1$. The squeeze view shows why $e^{x\ln x}$ has nowhere to go but $1$.)

Idea: peel the expression into (a piece that has a known limit) $\times$ (a bounded piece). $\frac{\sin (x^2)\cos (1/x)}{x}=\underset{\to 1}{\underbrace{\frac{\sin (x^2)}{x^2}}}\cdot \underset{\to 0}{\underbrace{x}}\cdot \underset{\text{bounded}}{\underbrace{\cos \frac{1}{x}}}.$ Why split like this: $\frac{\sin t}{t}\to 1$ is the standard trig limit (Limits of trigonometric functions) with $t=x^2\to 0$; the leftover $x$ vanishes and $\cos (1/x)$ stays inside $[-1,1]$.

Step 1 — regroup as (vanishing)$\times$(bounded). Let $B(x)=\frac{\sin (x^2)}{x^2}\cos \frac{1}{x}$. Since $\frac{\sin (x^2)}{x^2}\to 1$ it stays near $1$ (say within $[0,2]$) close to $0$, and $\cos (1/x)\in [-1,1]$, so $B(x)$ is bounded: $|B(x)|\le 2$ near $0$.

Step 2 — squeeze the whole thing. The expression equals $x\cdot B(x)$, and $-2|x|\le xB(x)\le 2|x|.$ Why $|x|$: multiplying $|B|\le 2$ by $|x|\ge 0$ (safe direction) bounds the product's size.

Step 3. Both walls $\pm 2|x|\to 0$, so ${\lim }_{x\to 0}\frac{\sin (x^2)\cos (1/x)}{x}=\boxed{0}.$

Step 1 — unfold the absolute value into two walls. $|f(x)-3|\le 5|x-2{|}^2$ means $3-5|x-2{|}^2\le f(x)\le 3+5|x-2{|}^2.$

Step 2 — squeeze. As $x\to 2$, $|x-2{|}^2\to 0$, so both walls $\to 3$. Hence ${\lim }_{x\to 2}f(x)=\boxed{3}.$

Step 3 — continuity. Since ${\lim }_{x\to 2}f(x)=3=f(2)$, by the definition in Continuity, $f$ is continuous at $2$. (This bound-and-squeeze is exactly how one proves differentiability-style estimates.)

Step 1 — bound the wiggle: $-1\le \sin \frac{1}{x}\le 1$.

Step 2 — multiply by $x$, but $x$ changes sign. This is the whole point.

• If $x>0$: multiplying preserves direction: $-x\le x\sin \frac{1}{x}\le x$.
• If $x<0$: multiplying by a negative flips directions: $-x\ge x\sin \frac{1}{x}\ge x$, i.e. $x\le x\sin \frac{1}{x}\le -x$.

In both cases the two walls are $x$ and $-x$ — just in swapped roles. The uniform way to write it (dodging the sign split) is with $|x|$: $-|x|\le x\sin \frac{1}{x}\le |x|,\text{true for all }x\ne 0.$ Why this is legit: $\left|x\sin \frac{1}{x}\right|=|x|\left|\sin \frac{1}{x}\right|\le |x|\cdot 1=|x|$.

Step 3 — squeeze. $\pm |x|\to 0$, so ${\lim }_{x\to 0}x\sin \frac{1}{x}=\boxed{0}.$ Look at the figure below: the graph lives entirely inside the "cone" $\pm |x|$ (chalk blue), touching neither wall's escape.

Conclusion is correct, but not because of the squeeze. The squeeze theorem can only ever produce a limit; it can never prove a limit fails. Walls that don't meet simply give no information — they don't forbid a limit.

The correct reasoning: compute the one-sided limits directly. For $x>0$, $\frac{|x|}{x}=\frac{x}{x}=1$; for $x<0$, $\frac{|x|}{x}=\frac{-x}{x}=-1$. Left limit $-1\ne 1$ right limit, so ${\lim }_{x\to 0}\frac{|x|}{x}\text{does not exist.}$ Answer: conclusion right, reasoning wrong — a non-meeting squeeze proves nothing.

Step 1 — lower wall. For $n\ge 1$, $n\ge 1\Rightarrow \sqrt[n]{n}\ge 1$. So the lower wall is the constant $1$.

Step 2 — build the upper wall. Write $\sqrt[n]{n}=1+t_n$ where $t_n\ge 0$. Raise to the $n$-th power: $n=(1+t_n{)}^n.$ By the binomial expansion, all terms are $\ge 0$, so keeping just the $k=2$ term (for $n\ge 2$): $n=(1+t_n{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2})t_n^2=\frac{n(n-1)}{2}{t}_n^2.$ Rearranging: $t_n^2\le \frac{2}{n-1}$, hence $0\le t_n\le \sqrt{\frac{2}{n-1}}$. Therefore the upper wall is $1\le \sqrt[n]{n}\le 1+\sqrt{\frac{2}{n-1}}.$

Step 3 — squeeze. As $n\to \infty$, $\sqrt{\frac{2}{n-1}}\to 0$, so the upper wall $\to 1$; lower wall is $1$. Both meet at $1$: ${\lim }_{n\to \infty }\sqrt[n]{n}=\boxed{1}.$ This is mastery-level because you manufactured the upper wall from an inequality no one gave you.

Do my two walls sandwich $f$ on a whole neighborhood of $a$? ::: Yes — inequality must hold near $a$, not only at $a$. Do both walls converge to the same number $L$? ::: Yes — different limits give no conclusion. When I multiplied/divided, was the factor guaranteed non-negative (or did I split by sign / use $|x|$)? ::: Yes — else the inequality could flip. Am I trying to prove a limit exists, not disprove one? ::: Squeeze only proves existence.