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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 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.
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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}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 21:15, 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 $\lim_{x\to\0}x^2 e^{\sin\frac{1}{x}}$ (Error compiling LaTeX. Unknown error_msg). 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. Unknown error_msg) is 0.


Template:Incomplete

See Also