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The Squeeze Theorem

April 8th, 2009 |

Author: Merrin

We mention the squeeze theorem because it is a useful way of evaluating certain trigonometric limits. These limits will be used in the next chapter as a logical way to evaluate the derivatives of sine and cosine. The squeeze theorem can be used to also evaluate the limits of some wildly oscillating functions if they are constrained by an envelope function which squeezes down to zero height. The squeeze theorem or the pinching theorem can be understood intuitively as follows. If three aligned points on the real line are constrained by moving the outer two points together, then eventually all three points meet together.

Theorem 1. The Squeeze Theorem states that if f(x)<g(x)<h(x) for all x near c then if

$\mathop{\lim }\limits_{x\to c} f(x)=\mathop{\lim }\limits_{x\to c} h(x)=L$

then

$\mathop{\lim }\limits_{x\to c} g(x)=L$

Example 1. Find the limit as x goes to zero of

$f(x) = x\sin (x^{-1} )$

Solution 1. The squeeze theorem can be used to find this limit. First note that the range of sine is between -1 and 1. We can write

$-1\le \sin x^{-1} \le 1$

Now we multiply everything by x and we find that

$-x\le x\sin x^{-1} \le x$

We take the limit of everything as x goes to zero and we find that

${\mathop{\lim }\limits_{x\to 0} (-x)\le \mathop{\lim }\limits_{x\to 0} (x\sin x^{-1} )\le \mathop{\lim }\limits_{x\to 0} (x)} \\ {0\le \mathop{\lim }\limits_{x\to 0} (x\sin x^{-1} )\le 0}$

By the squeeze theorem,

$\mathop{\lim }\limits_{x\to 0} (x\sin x^{-1} )=0$

We can also see this limit by plotting f(x) against the envelope functions y = x and y = -x. A similar limit follows for

$\mathop{\lim }\limits_{x\to 0} x^{n} \sin (x^{-1} )=0$

Except for n = 0 in which case

$\mathop{\lim }\limits_{x\to 0} \sin (x^{-1} )=DNE$

xsinxinv

The plot of the function xsin(1/x) reveals the limit at x = 0.

Example 2. Calculate the following limit using the squeeze theorem.

$\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } =1$

Calculating trigonometric limits by the squeeze theorem

squeezetrig

Solution 2. Consider three regions related to the angle theta: Delta ABC with base of cosine theta and an altitude sine theta, sector ADC with an area of theta/2 , and Delta ADE which has base 1 and a height of tangent theta. There is a hierarchy of the areas that can be represented as a compound inequality.

${\cfrac{1}{2} \cos \theta \sin \theta \le \cfrac{1}{2} \theta \le \cfrac{1}{2} \tan \theta } \\ {\cos \theta \le \cfrac{\theta }{\sin \theta } \le \cfrac{1}{\cos \theta } } \\ {\sec \theta \ge \cfrac{\sin \theta }{\theta } \ge \cos \theta }$

We now use the squeeze theorem and take the limit as theta goes to zero.

${\mathop{\lim }\limits_{\theta \to 0} \sec \theta \ge \mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } \ge \mathop{\lim }\limits_{\theta \to 0} \cos \theta }$

${1\, \ge \, \mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } \ge 1}$

Since cosine theta and secant theta both tend to one as theta goes to zero, we find the limit involving

$\cfrac{\sin \theta }{\theta }$

by the squeeze theorem.

$\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } =1$

Example 3. Given that

$\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } =1 \text{ find } \mathop{\lim }\limits_{\theta \to 0} \cfrac{1-\cos \theta }{\theta }$

Solution 3. We first manipulate the argument of limit to use the given formula.

${L}={\mathop{\lim }\limits_{\theta \to 0} \cfrac{\cos \theta -1}{\theta } =\mathop{\lim }\limits_{\theta \to 0} \cfrac{\cos \theta -1}{\theta } \cfrac{1+\cos \theta }{1+\cos \theta } } \\ = {\mathop{\lim }\limits_{\theta \to 0} \cfrac{\cos ^{2} \theta -1}{\theta (1+\cos \theta )} =-\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin ^{2} \theta }{\theta } \cfrac{1}{1+\cos \theta } } \\ ={-\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin ^{2} \theta }{\theta ^{2} } \cfrac{\theta }{1+\cos \theta } =-\left(\mathop{\lim }\limits_{\theta \to 0} \cfrac{\sin \theta }{\theta } \right)^{2} \mathop{\lim }\limits_{\theta \to 0} \cfrac{\theta }{1+\cos \theta } =-1^{2} \cdot \cfrac{0}{2} =0}$

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The Squeeze Theorem

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