Editing Squeeze theorem (section)

== Examples ==

=== First example ===

[[File:Inst_satsen.svg|thumb|right|250px|<math>x^2 \sin\left(\tfrac{1}{x}\right)</math> being squeezed in the limit as {{mvar|x}} goes to 0]]

The limit

<math display="block">\lim_{x \to 0}x^2 \sin\left( \tfrac{1}{x} \right)</math>

cannot be determined through the limit law

<math display="block">\lim_{x \to a}(f(x) \cdot g(x)) = 
\lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x),</math>

because

<math display="block">\lim_{x\to 0}\sin\left( \tfrac{1}{x} \right)</math>

does not exist.

However, by the definition of the [[sine function]],

<math display="block">-1 \le \sin\left( \tfrac{1}{x} \right) \le 1. </math>

It follows that

<math display="block">-x^2 \le x^2 \sin\left( \tfrac{1}{x} \right) \le x^2 </math>

Since <math>\lim_{x\to 0}-x^2 = \lim_{x\to 0}x^2 = 0</math>, by the squeeze theorem, <math>\lim_{x\to 0} x^2 \sin\left(\tfrac{1}{x}\right)</math> must also be 0.

=== Second example ===
[[File:Limit_sin_x_x.svg|thumb|upright=1.5|Comparing areas:<br/>
<math>\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}</math>]] 

Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities
<math display="block">
\begin{align}
& \lim_{x\to 0} \frac{\sin x}{x} =1, \\[10pt]
& \lim_{x\to 0} \frac{1 - \cos x}{x} = 0.
\end{align}
</math>

The first limit follows by means of the squeeze theorem from the fact that<ref>Selim 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]])</ref>

<math display="block"> \cos x \leq \frac{\sin x}{x} \leq 1 </math>

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

<math display="block"> 0 \leq  \frac{1 - \cos x}{x}  \leq  x </math>
for {{mvar|x}} close enough to 0. This can be derived by replacing {{math|sin ''x''}} in the earlier fact by <math display="inline"> \sqrt{1-\cos^2 x}</math> 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
<math display="block"> \frac{d}{d\theta} \tan\theta = \sec^2\theta </math>
by squeezing, as follows.

[[File:Tangent.squeeze.svg|thumb|upright=1.5|right]]

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

<math display="block"> \frac{\sec^2\theta\,\Delta\theta}{2}, </math>

since the radius is {{math|sec ''θ''}} and the arc on the [[unit circle]] has length&nbsp;{{math|Δ''θ''}}.  Similarly, the area of the larger of the two shaded sectors is

<math display="block"> \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}. </math>

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'' + Δ''θ'') − tan ''θ''}}, and the height is&nbsp;1. The area of the triangle is therefore

<math display="block"> \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2}. </math>

From the inequalities

<math display="block"> \frac{\sec^2\theta\,\Delta\theta}{2} \le \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2} \le \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2} </math>

we deduce that

<math display="block"> \sec^2\theta \le \frac{\tan(\theta + \Delta\theta) - \tan\theta}{\Delta\theta} \le \sec^2(\theta + \Delta\theta),</math>

provided&nbsp;{{math|Δ''θ'' > 0}}, and the inequalities are reversed if&nbsp;{{math|Δ''θ'' < 0}}.  Since the first and third expressions approach {{math|sec<sup>2</sup>''θ''}} as {{math|Δ''θ'' → 0}}, and the middle expression approaches <math>\tfrac{d}{d\theta} \tan\theta,</math> 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.<ref>{{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}}</ref>

<math display="block">\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2}</math>

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

<math display="block">\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}</math> 

therefore, by the squeeze theorem,

<math display="block">\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} = 0.</math>