Proof

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

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

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

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

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

As

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

means that

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

(1)

and

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

means that

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

(2)

then we have

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

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

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

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

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

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.

First example

${\displaystyle x^{2}\sin \left({\tfrac {1}{x}}\right)}$ being squeezed in the limit as x goes to 0

The limit

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

cannot be determined through the limit law

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

because

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

does not exist.

However, by the definition of the sine function,

$${\displaystyle -1\leq \sin \left({\tfrac {1}{x}}\right)\leq 1.}$$

It follows that

$${\displaystyle -x^{2}\leq x^{2}\sin \left({\tfrac {1}{x}}\right)\leq x^{2}}$$

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

Second example

Comparing areas:
${\displaystyle {\begin{array}{cccccc}&A(\triangle ADB)&\leq &A({\text{sector }}ADB)&\leq &A(\triangle ADF)\\[4pt]\Rightarrow &{\frac {1}{2}}\cdot \sin x\cdot 1&\leq &{\frac {x}{2\pi }}\cdot \pi &\leq &{\frac {1}{2}}\cdot \tan x\cdot 1\\[4pt]\Rightarrow &\sin x&\leq &x&\leq &{\frac {\sin x}{\cos x}}\\[4pt]\Rightarrow &{\frac {\cos x}{\sin x}}&\leq &{\frac {1}{x}}&\leq &{\frac {1}{\sin x}}\\[4pt]\Rightarrow &\cos x&\leq &{\frac {\sin x}{x}}&\leq &1\end{array}}}$

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

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

The first limit follows by means of the squeeze theorem from the fact that^[2]

$${\displaystyle \cos x\leq {\frac {\sin x}{x}}\leq 1}$$

for x close enough to 0. The correctness of which for positive x can be seen by simple geometric reasoning (see drawing) that can be extended to negative x as well. The second limit follows from the squeeze theorem and the fact that

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

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

These two 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 example

It 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

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

since the radius is sec θ and the arc on the unit circle has length Δθ. Similarly, the area of the larger of the two 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 height is 1. The area of the triangle is therefore

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

From the inequalities

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

we deduce that

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

provided Δθ > 0, and the inequalities are reversed if Δθ < 0. Since the first and third expressions approach sec²θ as Δθ → 0, and the middle expression approaches ${\displaystyle {\tfrac {d}{d\theta }}\tan \theta ,}$ the desired result follows.

Fourth example

The squeeze theorem can still be used in multivariable calculus but the lower (and upper functions) must be below (and above) the target function not just along a path but around the entire neighborhood of the point of interest and it only works if the function really does have a limit there. It can, therefore, be used to prove that a function has a limit at a point, but it can never be used to prove that a function does not have a limit at a point.^[3]

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

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

$${\displaystyle {\begin{array}{rccccc}&0&\leq &\displaystyle {\frac {x^{2}}{x^{2}+y^{2}}}&\leq &1\\[4pt]-|y|\leq y\leq |y|\implies &-|y|&\leq &\displaystyle {\frac {x^{2}y}{x^{2}+y^{2}}}&\leq &|y|\\[4pt]{{\displaystyle \lim _{(x,y)\to (0,0)}-|y|=0} \atop {\displaystyle \lim _{(x,y)\to (0,0)}\ \ \ |y|=0}}\implies &0&\leq &\displaystyle \lim _{(x,y)\to (0,0)}{\frac {x^{2}y}{x^{2}+y^{2}}}&\leq &0\end{array}}}$$

therefore, by the squeeze theorem,

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

Notes

References