Squeeze Theorem

Suppose in a particular mathematical situation there are two continuous functions $u(t)$u(t) and $v(t)$v(t).  It is found that as $t$t approaches a certain value $t_0$t0​, both functions approach the same limiting value $L$L.  If there is another continuous function $y(t)$y(t) that, in the region near $t_0$t0​, takes values that are always between the values of $u$u and $v$v, then $y$y also must approach the limit $L$L. In a sense it is squeezed between the values of the two functions $u$u and $v$v.

While this theorem seems intuitively obvious, it does have a rigorous proof based on the formal definition of a limit. (We will rely on intuition, for the moment.)

 

Example 

In the calculus of trigonometric functions, we wish to establish the derivative of the sine function. According to first principles in differentiation, if $f(x)=\sin x$f(x)=sinx then the derivative is

$f'(x)=\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$f′(x)=limh→0sin(x+h)−sinxh​

To simplify this limit, we expand the term $\sin(x+h)$sin(x+h) and do some rearranging.

$f'(x)$f′(x)	$=$=	$\lim_{h\rightarrow0}\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}$limh→0sinxcosh+sinhcosx−sinxh​
	$=$=	$\lim_{h\rightarrow0}\frac{\sin x\left(\cos h-1\right)+\sin h\cos x}{h}$limh→0sinx(cosh−1)+sinhcosxh​
	$=$=	$\lim_{h\rightarrow0}\frac{\cos h-1}{h}\sin x+\lim_{h\rightarrow0}\frac{\sin h}{h}\cos x$limh→0cosh−1h​sinx+limh→0sinhh​cosx

 

It appears that to make any headway, we need to evaluate the limits $\lim_{h\rightarrow0}\frac{\cos h-1}{h}$limh→0cosh−1h​ and $\lim_{h\rightarrow0}\frac{\sin h}{h}$limh→0sinhh​. Consider the following diagram. We make an argument based on $\theta$θ in the first quadrant but it is equally true in the fourth quadrant.

 

The height of triangle $0AC$0AC is $\sin\theta$sinθ. Therefore the area of this triangle is $\frac{1}{2}\sin\theta$12​sinθ.

The sector $0AD$0AD has area $\frac{1}{2}\theta$12​θ      ($=\frac{\theta}{2\pi}\times1^2\pi$=θ2π​×12π).

The height of triangle $0BD$0BD is $\tan\theta$tanθ. So, Triangle $0BD$0BD has area $\frac{1}{2}\tan\theta$12​tanθ.

Now, because of the way these shapes are nested within one another, we can compare the areas and write

$\frac{1}{2}\sin\theta\le\frac{1}{2}\theta\le\frac{1}{2}\tan\theta$12​sinθ≤12​θ≤12​tanθ

and therefore,

$\sin\theta\le\theta\le\frac{\sin\theta}{\cos\theta}$sinθ≤θ≤sinθcosθ​.

Taking the reciprocals of each term, we have

$\frac{1}{\sin\theta}\ge\frac{1}{\theta}\ge\frac{\cos\theta}{\sin\theta}$1sinθ​≥1θ​≥cosθsinθ​

and finally, since $\sin\theta$sinθ is positive in the first quadrant,

$1\ge\frac{\sin\theta}{\theta}\ge\cos\theta$1≥sinθθ​≥cosθ

We are now in a position to apply the squeezing theorem to obtain $\lim_{h\rightarrow0}\frac{\sin h}{h}$limh→0sinhh​.

Since $\lim_{h\rightarrow0}1=1$limh→01=1 and $\lim_{h\rightarrow0}\cos h=1$limh→0cosh=1, we conclude that 

$\lim_{h\rightarrow0}\frac{\sin h}{h}=1$limh→0sinhh​=1.

The other limit, $\lim_{h\rightarrow0}\frac{\cos h-1}{h}$limh→0cosh−1h​ is found as follows.

$\lim_{h\rightarrow0}\frac{\cos h-1}{h}$limh→0cosh−1h​	$=$=	$\lim_{h\rightarrow0}\left[\frac{\cos h-1}{h}\times\frac{\cos h+1}{\cos h+1}\right]$limh→0[cosh−1h​×cosh+1cosh+1​]
	$=$=	$\lim_{h\rightarrow0}\frac{\cos^2h-1}{h(\cos h+1)}$limh→0cos2h−1h(cosh+1)​
	$=$=	$\lim_{h\rightarrow0}\frac{-\sin^2h}{h(\cos h+1)}$limh→0−sin2hh(cosh+1)​
	$=$=	$\lim_{h\rightarrow0}\frac{\sin h}{h}\times\lim_{h\rightarrow0}\frac{-\sin h}{\cos h+1}$limh→0sinhh​×limh→0−sinhcosh+1​
	$=$=	$1\times0$1×0
	$=$=	$0$0

 

We can now write the derivative of $f(x)=\sin x$f(x)=sinx. It is, $f'(x)=\cos x$f′(x)=cosx.