Easy Way to Understand the Squeeze Theorem

On calculating limits past bounding a function between two other functions

"Sandwich theorem" redirects hither. For the result in measure theory, see Ham sandwich theorem.

Analogy of the squeeze theorem

When a sequence lies between 2 other converging sequences with the same limit, it also converges to this limit.

In calculus, the squeeze theorem (also known as the sandwich theorem, among other names^[a]) is a theorem regarding the limit of a function that is trapped between two other functions.

The clasp theorem is used in calculus and mathematical analysis, typically to confirm the limit of a function via comparison with ii other functions whose limits are known. It was first used geometrically by the mathematicians Archimedes and Eudoxus in an endeavour to compute π, and was formulated in modernistic terms past Carl Friedrich Gauss.

In many languages (e.m. French, German, Italian, Hungarian and Russian), the squeeze theorem is also known as the two officers (and a boozer) theorem, or some variation thereof.^{[ commendation needed ]} The story is that if two police officers are escorting a drunk prisoner betwixt them, and both officers go to a cell, so (regardless of the path taken, and the fact that the prisoner may be wobbling most between the officers) the prisoner must also finish up in the jail cell.

Statement [edit]

The clasp theorem is formally stated as follows.^[ane]

Theorem  — Let I exist an interval containing the point a. Let one thousand, f, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have

$${\displaystyle thou(x)\leq f(x)\leq h(x)}$$

and too suppose that

$${\displaystyle \lim _{x\to a}k(x)=\lim _{x\to a}h(x)=Fifty.}$$

And so ${\displaystyle \lim _{x\to a}f(ten)=Fifty.}$

This theorem is too valid for sequences. Let ${\displaystyle (a_{north}),(c_{due north})}$ exist ii sequences converging to ${\displaystyle \ell }$ , and ${\displaystyle (b_{northward})}$ a sequence. If ${\displaystyle \forall due north\geq Due north,N\in \mathbb {Due north} }$ we have ${\displaystyle a_{due north}\leq b_{due north}\leq c_{n}}$ , then ${\displaystyle (b_{n})}$ also converges to ${\displaystyle \ell }$ .

Proof [edit]

According to the higher up hypotheses we have, taking the limit inferior and superior:

$${\displaystyle L=\lim _{x\to a}thou(x)\leq \liminf _{ten\to a}f(x)\leq \limsup _{x\to a}f(x)\leq \lim _{10\to a}h(x)=L,}$$

and then all the inequalities are indeed equalities, and the thesis immediately follows.

A direct proof, using the ${\displaystyle (\varepsilon ,\delta )}$ -definition of limit, would be to prove that for all real ${\textstyle \varepsilon >0}$ there exists a real ${\displaystyle \delta >0}$ such that for all ${\displaystyle x}$ with ${\displaystyle |ten-a|<\delta }$ , we have ${\displaystyle |f(x)-L|<\varepsilon }$ . Symbolically,

$${\displaystyle \forall \varepsilon >0,\exists \delta >0:\forall x,(|x-a|<\delta \ \Rightarrow |f(10)-L|<\varepsilon ).}$$

As

$${\displaystyle \lim _{x\to a}thou(10)=L}$$

means that

$${\displaystyle \forall \varepsilon >0,\exists \ \delta _{one}>0:\forall ten\ (|x-a|<\delta _{1}\ \Rightarrow \ |g(x)-L|<\varepsilon ).}$$

(1)

and

$${\displaystyle \lim _{x\to a}h(10)=Fifty}$$

means that

$${\displaystyle \forall \varepsilon >0,\exists \ \delta _{2}>0:\forall x\ (|x-a|<\delta _{2}\ \Rightarrow \ |h(ten)-L|<\varepsilon ),}$$

(2)

and so we have

$${\displaystyle g(10)\leq f(x)\leq h(10)}$$

$${\displaystyle g(x)-L\leq f(ten)-L\leq h(x)-L}$$

We can choose ${\displaystyle \delta :=\min \left\{\delta _{1},\delta _{2}\right\}}$ . And then, if ${\displaystyle |x-a|<\delta }$ , combining (1) and (2), we have

${\displaystyle -\varepsilon <g(x)-L\leq f(10)-L\leq h(x)-L\ <\varepsilon ,}$

$${\displaystyle -\varepsilon <f(x)-L<\varepsilon ,}$$

which completes the proof. Q.E.D

The proof for sequences is very similar, using the ${\displaystyle \varepsilon }$ -definition of the limit of a sequence.

Examples [edit]

First example [edit]

x ² sin(1/x) being squeezed in the limit as x goes to 0

The limit

$${\displaystyle \lim _{x\to 0}x^{2}\sin({\tfrac {1}{x}})}$$

cannot be adamant through the limit law

$${\displaystyle \lim _{x\to a}(f(10)\cdot g(x))=\lim _{ten\to a}f(x)\cdot \lim _{x\to a}g(x),}$$

because

$${\displaystyle \lim _{x\to 0}\sin({\tfrac {1}{x}})}$$

does not exist.

However, by the definition of the sine function,

$${\displaystyle -one\leq \sin({\tfrac {1}{ten}})\leq one.}$$

It follows that

$${\displaystyle -x^{2}\leq ten^{2}\sin({\tfrac {one}{ten}})\leq x^{2}}$$

Since ${\displaystyle \lim _{x\to 0}-10^{2}=\lim _{ten\to 0}ten^{2}=0}$ , past the squeeze theorem, ${\displaystyle \lim _{10\to 0}x^{2}\sin({\tfrac {1}{x}})}$ must also be 0.

Second example [edit]

Comparison areas:
${\displaystyle {\begin{aligned}&\,A(\triangle ADF)\geq A({\text{sector}}\,ADB)\geq A(\triangle ADB)\\\Rightarrow &\,{\frac {1}{2}}\cdot \tan(ten)\cdot 1\geq {\frac {x}{ii\pi }}\cdot \pi \geq {\frac {one}{2}}\cdot \sin(x)\cdot ane\\\Rightarrow &\,{\frac {\sin(x)}{\cos(x)}}\geq ten\geq \sin(x)\\\Rightarrow &\,{\frac {\cos(10)}{\sin(x)}}\leq {\frac {1}{x}}\leq {\frac {1}{\sin(x)}}\\\Rightarrow &\,\cos(x)\leq {\frac {\sin(10)}{x}}\leq i\end{aligned}}}$

Probably the all-time-known examples of finding a limit by squeezing are the proofs of the equalities

$${\displaystyle {\brainstorm{aligned}&\lim _{ten\to 0}{\frac {\sin(ten)}{x}}=1,\\[10pt]&\lim _{ten\to 0}{\frac {one-\cos(ten)}{x}}=0.\end{aligned}}}$$

The first limit follows past ways of the squeeze theorem from the fact that^[two]

$${\displaystyle \cos 10\leq {\frac {\sin(x)}{10}}\leq 1}$$

for x close enough to 0. The definiteness of which for positive x tin exist seen by simple geometric reasoning (see drawing) that can be extended to negative 10 likewise. The 2nd limit follows from the clasp theorem and the fact that

$${\displaystyle 0\leq {\frac {1-\cos(10)}{x}}\leq x}$$

for x shut enough to 0. This can be derived by replacing ${\displaystyle \sin(x)}$ in the before fact by ${\textstyle {\sqrt {1-\cos ^{2}(x)}}}$ and squaring the resulting inequality.

These ii limits are used in proofs of the fact that the derivative of the sine function is the cosine function. That fact is relied on in other proofs of derivatives of trigonometric functions.

Third case [edit]

Information technology is possible to show that

$${\displaystyle {\frac {d}{d\theta }}\tan \theta =\sec ^{2}\theta }$$

by squeezing, as follows.

In the illustration at right, the area of the smaller of the two shaded sectors of the circle is

$${\displaystyle {\frac {\sec ^{two}\theta \,\Delta \theta }{2}},}$$

since the radius is secθ and the arc on the unit circumvolve has length Δθ. Similarly, the expanse of the larger of the 2 shaded sectors is

$${\displaystyle {\frac {\sec ^{2}(\theta +\Delta \theta )\,\Delta \theta }{2}}.}$$

What is squeezed between them is the triangle whose base is the vertical segment whose endpoints are the two dots. The length of the base of the triangle is tan(θ + Δθ) − tan(θ), and the acme is 1. The area of the triangle is therefore

$${\displaystyle {\frac {\tan(\theta +\Delta \theta )-\tan(\theta )}{2}}.}$$

From the inequalities

$${\displaystyle {\frac {\sec ^{two}\theta \,\Delta \theta }{ii}}\leq {\frac {\tan(\theta +\Delta \theta )-\tan(\theta )}{2}}\leq {\frac {\sec ^{2}(\theta +\Delta \theta )\,\Delta \theta }{2}}}$$

we deduce that

$${\displaystyle \sec ^{ii}\theta \leq {\frac {\tan(\theta +\Delta \theta )-\tan(\theta )}{\Delta \theta }}\leq \sec ^{2}(\theta +\Delta \theta ),}$$

provided Δθ > 0, and the inequalities are reversed if Δθ < 0. Since the first and 3rd expressions approach sec² θ equally Δθ → 0, and the center expression approaches d / dθ  tanθ, the desired result follows.

Fourth example [edit]

The clasp theorem tin still be used in multivariable calculus only the lower (and upper functions) must be below (and above) the target function not only along a path but around the entire neighborhood of the betoken of interest and it but works if the part actually does take a limit at that place. It tin can, therefore, be used to prove that a role has a limit at a point, but it can never be used to prove that a part does not accept a limit at a point.^[3]

$${\displaystyle \lim _{(ten,y)\to (0,0)}{\frac {x^{2}y}{10^{2}+y^{ii}}}}$$

cannot be found by taking any number of limits forth paths that pass through the point, only since

$${\displaystyle 0\leq {\frac {ten^{2}}{x^{2}+y^{two}}}\leq 1}$$

$${\displaystyle -\left|y\correct\vert \leq y\leq \left|y\correct\vert }$$

$${\displaystyle -\left|y\correct\vert \leq {\frac {x^{two}y}{x^{2}+y^{2}}}\leq \left|y\right\vert }$$

$${\displaystyle \lim _{(10,y)\to (0,0)}-\left|y\right\vert =0}$$

$${\displaystyle \lim _{(x,y)\to (0,0)}\left|y\right\vert =0}$$

$${\displaystyle 0\leq \lim _{(x,y)\to (0,0)}{\frac {x^{2}y}{10^{2}+y^{2}}}\leq 0}$$

therefore, by the clasp theorem,

$${\displaystyle \lim _{(x,y)\to (0,0)}{\frac {10^{two}y}{x^{2}+y^{ii}}}=0}$$

References [edit]

Notes [edit]

1. ^ Too known as the pinching theorem, the sandwich dominion, the police theorem, the between theorem and sometimes the squeeze lemma. In Italian republic, the theorem is as well known equally the theorem of carabinieri.

References [edit]

1. ^ Sohrab, Houshang H. (2003). Basic Existent Assay (2nd ed.). Birkhäuser. p. 104. ISBN978-1-4939-1840-9.
2. ^ Selim G. Krejn, V.N. Uschakowa: Vorstufe zur höheren Mathematik. Springer, 2013, ISBN 9783322986283, pp. 80-81 (German). See also Sal Khan: Proof: limit of (sin 10)/x at x=0 (video, Khan Academy)
3. ^ Stewart, James (2008). "Chapter xv.2 Limits and Continuity". Multivariable Calculus (6th ed.). pp. 909–910. ISBN978-0495011637.

External links [edit]

• Weisstein, Eric Westward. "Squeezing Theorem". MathWorld.
• Squeeze Theorem by Bruce Atwood (Beloit College) after work past, Selwyn Hollis (Armstrong Atlantic State University), the Wolfram Demonstrations Project.
• Squeeze Theorem on ProofWiki.

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Source: https://en.wikipedia.org/wiki/Squeeze_theorem