Squeeze theorem#Second example

The squeeze theorem is formally stated as follows.{{cite book|last1=Sohrab|first1=Houshang H.|title=Basic Real Analysis| date=2003|publisher=Birkhäuser|isbn=978-1-4939-1840-9|page=104|edition=2nd|url=https://books.google.com/books?id=QnpqBQAAQBAJ&pg=PA104}}

{{math theorem|

Let {{mvar|I}} be an interval containing the point {{mvar|a}}. Let {{mvar|g}}, {{mvar|f}}, and {{mvar|h}} be functions defined on {{mvar|I}}, except possibly at {{mvar|a}} itself. Suppose that for every {{mvar|x}} in {{mvar|I}} not equal to {{mvar|a}}, we have

$$g(x)\; \backslash leq\; f(x)\; \backslash leq\; h(x)$$

and also suppose that

$$\backslash lim\_\{x\; \backslash to\; a\}\; g(x)\; =\; \backslash lim\_\{x\; \backslash to\; a\}\; h(x)\; =\; L.$$

Then $\backslash lim\_\{x\; \backslash to\; a\}\; f(x)\; =\; L.$

}}

• The functions {{mvar|g}} and {{mvar|h}} are said to be lower and upper bounds (respectively) of {{mvar|f}}.
• Here, {{mvar|a}} is not required to lie in the interior of {{mvar|I}}. Indeed, if {{mvar|a}} is an endpoint of {{mvar|I}}, then the above limits are left- or right-hand limits.
• A similar statement holds for infinite intervals: for example, if {{math|1=I = (0, ∞)}}, then the conclusion holds, taking the limits as {{math|x → ∞}}.

This theorem is also valid for sequences. Let {{math|(a{{sub|n}}), (c{{sub|n}})}} be two sequences converging to {{mvar|ℓ}}, and {{math|(b{{sub|n}})}} a sequence. If $\backslash forall\; n\backslash geq\; N,\; N\backslash in\backslash N$ we have {{math|a{{sub|n}} ≤ b{{sub|n}} ≤ c{{sub|n}}}}, then {{math|(b{{sub|n}})}} also converges to {{mvar|ℓ}}.

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

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

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

A direct proof, using the {{math|(ε, δ)}}-definition of limit, would be to prove that for all real {{math|ε > 0}} there exists a real {{math|δ > 0}} such that for all {{mvar|x}} with we have Symbolically,

As

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

means that

{{NumBlk|||{{EquationRef|1}}}}

and

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

means that

{{NumBlk|||{{EquationRef|2}}}}

then we have

$$g(x)\; \backslash leq\; f(x)\; \backslash leq\; h(x)$$

$$g(x)\; -\; L\backslash leq\; f(x)\; -\; L\backslash leq\; h(x)\; -\; L$$

We can choose $\backslash delta:=\backslash min\backslash left\backslash \{\backslash delta\_1,\backslash delta\_2\backslash right\backslash \}$. Then, if , combining ({{EquationNote|1}}) and ({{EquationNote|2}}), we have

which completes the proof. Q.E.D

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

File:Inst_satsen.svg

The limit

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

cannot be determined through the limit law

$$\backslash lim\_\{x\; \backslash to\; a\}(f(x)\; \backslash cdot\; g(x))\; =$$

\lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x),

because

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

does not exist.

However, by the definition of the sine function,

$$-1\; \backslash le\; \backslash sin\backslash left(\; \backslash tfrac\{1\}\{x\}\; \backslash right)\; \backslash le\; 1.$$

It follows that

$$-x^2\; \backslash le\; x^2\; \backslash sin\backslash left(\; \backslash tfrac\{1\}\{x\}\; \backslash right)\; \backslash le\; x^2$$

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

[[File:Limit_sin_x_x.svg|thumb|upright=1.5|Comparing areas:

$\backslash begin\{array\}\{cccccc\}$

& 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

\begin{align}

& \lim_{x\to 0} \frac{\sin x}{x} =1, \\[10pt]

& \lim_{x\to 0} \frac{1 - \cos x}{x} = 0.

\end{align}

The first limit follows by means of the squeeze theorem from the fact thatSelim G. Krejn, V.N. Uschakowa: Vorstufe zur höheren Mathematik. Springer, 2013, {{ISBN|9783322986283}}, pp. [https://books.google.com/books?id=-yXMBgAAQBAJ&pg=PA80 80-81] (German). See also Sal Khan: [https://www.khanacademy.org/math/ap-calculus-ab/limits-from-equations-ab/squeeze-theorem-ab/v/proof-lim-sin-x-x Proof: limit of (sin x)/x at x=0] (video, Khan Academy)

$$\backslash cos\; x\; \backslash leq\; \backslash frac\{\backslash sin\; x\}\{x\}\; \backslash leq\; 1$$

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

$$0\; \backslash leq\; \backslash frac\{1\; -\; \backslash cos\; x\}\{x\}\; \backslash leq\; x$$

for {{mvar|x}} close enough to 0. This can be derived by replacing {{math|sin x}} in the earlier fact by $\backslash sqrt\{1-\backslash cos^2\; x\}$ 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.

It is possible to show that

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

by squeezing, as follows.

File:Tangent.squeeze.svg

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

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

since the radius is {{math|sec θ}} and the arc on the unit circle has length {{math|Δθ}}. Similarly, the area of the larger of the two shaded sectors is

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

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 {{math|tan(θ + Δθ) − tan θ}}, and the height is 1. The area of the triangle is therefore

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

From the inequalities

$$\backslash frac\{\backslash sec^2\backslash theta\backslash ,\backslash Delta\backslash theta\}\{2\}\; \backslash le\; \backslash frac\{\backslash tan(\backslash theta\; +\; \backslash Delta\backslash theta)\; -\; \backslash tan\backslash theta\}\{2\}\; \backslash le\; \backslash frac\{\backslash sec^2(\backslash theta\; +\; \backslash Delta\backslash theta)\backslash ,\backslash Delta\backslash theta\}\{2\}$$

we deduce that

$$\backslash sec^2\backslash theta\; \backslash le\; \backslash frac\{\backslash tan(\backslash theta\; +\; \backslash Delta\backslash theta)\; -\; \backslash tan\backslash theta\}\{\backslash Delta\backslash theta\}\; \backslash le\; \backslash sec^2(\backslash theta\; +\; \backslash Delta\backslash theta),$$

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

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.{{cite book|chapter=Chapter 15.2 Limits and Continuity| pages=909–910|title=Multivariable Calculus|year=2008|last1=Stewart|first1=James| author-link1=James Stewart (mathematician)| edition=6th|isbn=978-0495011637}}

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

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

$$\backslash begin\{array\}\{rccccc\}$$

& 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,

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

{{notelist}}

• {{MathWorld |title=Squeeze Theorem |urlname=SqueezeTheorem}}
• [http://demonstrations.wolfram.com/SqueezeTheorem/ Squeeze Theorem] by Bruce Atwood (Beloit College) after work by, Selwyn Hollis (Armstrong Atlantic State University), the Wolfram Demonstrations Project.
• [https://proofwiki.org/wiki/Squeeze_Theorem Squeeze Theorem] on ProofWiki.

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