Difference between revisions of "Squeeze Theorem"
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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 | + | 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.
Contents
[hide]Theorem
Suppose is between and for all in a neighborhood of the point . If and approach some common limit as approaches , then .
Proof
If is between and for all in the neighborhood of , then either or for all in this neighborhood. The two cases are the same up to renaming our functions, so assume without loss of generality that .
We must show that for all there is some for which implies .
Now since , there must exist such that
Now let . If then
So . Now by the definition of a limit we get 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 with thelimit $\lim_{x\to\0} f(x)=x^2 e^{\sin\frac{1}{x}}$ (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, and . 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.