Difference between revisions of "Squeeze Theorem"

(Applications and examples)
(Applications and examples)
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== Applications and examples==
 
== Applications and examples==
The Squeeze Theorem can be used to evaluate limits that might not normally be defined. An example is the limit <math>\lim_{x\to\0} f(x)=x^2e^{\sin\frac{1}{x}}</math>. The limit is not normally defined, because the function oscillates infinitely many times around 0, but it can be evaluated with the Squeeze Theorem as following. Create two functions, <math>x^2</math> and <math>-x^2</math>. It is easy to see that around 0, the function in question is squeezed between these two functions, and the limit as both of these approach 0 is 0, so <math>\lim_{x\to\0} x^2e^{\sin\frac{1}{x}}</math> is 0.
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The Squeeze Theorem can be used to evaluate limits that might not normally be defined. An example is the function <math>f(x)=x^2 e^{\sin\frac{1}{x}}</math>  with thelimit <math>\lim_{x\to\0} f(x)=x^2 e^{\sin\frac{1}{x}}</math>. The limit is not normally defined, because the function oscillates infinitely many times around 0, but it can be evaluated with the Squeeze Theorem as following. Create two functions, <math>x^2</math> and <math>-x^2</math>. It is easy to see that around 0, the function in question is squeezed between these two functions, and the limit as both of these approach 0 is 0, so <math>\lim_{x\to\0} x^2 e^{\sin\frac{1}{x}}</math> is 0.
  
  

Revision as of 21:14, 28 August 2015

The Squeeze Theorem (also called the Sandwich Theorem or the Squeeze Play Theorem) is a relatively simple theorem that deals with calculus, specifically limits.

Squeeze Theorem

Theorem

Suppose $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in a neighborhood of the point $S$. If $g$ and $h$ approach some common limit $L$ as $x$ approaches $S$, then $\lim_{x\to S}f(x)=L$.

Proof

If $f(x)$ is between $g(x)$ and $h(x)$ for all $x$ in the neighborhood of $S$, then either $g(x)\leq f(x) \leq h(x)$ or $h(x)\leq f(x)\leq g(x)$ for all $x$ in this neighborhood. The two cases are the same up to renaming our functions, so assume without loss of generality that $g(x)\leq f(x) \leq h(x)$.

We must show that for all $\varepsilon >0$ there is some $\delta > 0$ for which $|x-S|<\delta$ implies $|f(x)-L|<\varepsilon$.

Now since $\lim_{x\to S}g(x)=\lim_{x\to S}h(x)=L$, there must exist $\delta_1,\delta_2>0$ such that

\[|x-S|<\delta_1 \Rightarrow |g(x)-L|<\varepsilon \textrm{  and  } |x-S|<\delta_2 \Rightarrow |h(x)-L|<\varepsilon.\]

Now let $\delta = \min\{\delta_1,\delta_2\}$. If $|x-S|<\delta$ then

$-\varepsilon < g(x) - L \leq f(x) - L \leq h(x) - L < \varepsilon.$

So $|f(x)-L|<\varepsilon$. Now by the definition of a limit we get $\lim_{x\to S}f(x)=L$ as desired.

Applications and examples

The Squeeze Theorem can be used to evaluate limits that might not normally be defined. An example is the function $f(x)=x^2 e^{\sin\frac{1}{x}}$ with thelimit $\lim_{x\to\0} f(x)=x^2 e^{\sin\frac{1}{x}}$ (Error compiling LaTeX. ! Undefined control sequence.). The limit is not normally defined, because the function oscillates infinitely many times around 0, but it can be evaluated with the Squeeze Theorem as following. Create two functions, $x^2$ and $-x^2$. It is easy to see that around 0, the function in question is squeezed between these two functions, and the limit as both of these approach 0 is 0, so $\lim_{x\to\0} x^2 e^{\sin\frac{1}{x}}$ (Error compiling LaTeX. ! Undefined control sequence.) is 0.


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See Also

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