Thursday, February 7, 2013

Squeeze Theorem Proof


The proof is a correct demonstration of math statement. It should be true. The proof of one statement is used in other statement proof. The limits of a function are defined by squeeze theorem. We can also refer the squeeze theorem as sandwich theorem or pinching theorem. Now we are going to see about squeeze theorem.

Explanation for Squeeze Theorem Proof

Define squeeze theorem:

In calculus, the squeeze theorem is used and we can analyze the function’s limits by using this theorem. The interval ‘I’ is including the point ‘a’. The functions f, g and h are derived by interval ‘I’. We gets the function as g(x) `lt=` f(x) `lt=` h(x) if we have unequal 'x' values present. In other words, the definition is `lim_(x->a)` g(x) = `lim_(x->a)` h(x) = L. Final result is `lim_(x->a)` f(x). The lower bound g(x) and upper bound h(x) are used to bounds the f(x). We does not including the value 'a' in the interval 'I'.

I like to share this Calculus Limits with you all through my article.

More about Squeeze Theorem Proof

The squeeze theorem proof is derived by previous statements and the proof of this theorem is using the special case and general notations.

Using the special cased for squeeze theorem proof:

The special case is g(x) = 0 for all x and L = 0.

First prove the special case as,

Let us take the special case as `lim_(x->a)` h(x) = 0.

Use the fixed positive number ε > 0 and δ > 0 from limits of function.

if 0 < | x-a | < δ then |h(x)| < ε.

Take the terms from above interval and the terms are 0 = g(x) `<=` f(x) `<=` h(x). So | f(x) | `<=` | h(x) |.

The conclusion of theorem is if 0 < | x-a | < δ then | f(x) | `<=` | h(x) | `<` ε.

Final proof of squeeze theorem by above condition is `lim_(x->a)` f(x) = 0 = L.

Use the general notations for squeeze theorem proof:

The g and L are arbitrary and these are used for proof. We have g(x) `<=` f(x) `<=` h(x).

Subtract the g(x) from both sides as 0 `<=` f(x) - g(x) `<=` h(x) - g(x).

Let us take x--> a and g(x) and h(x) as 'L'.

Therefore, h(x) - g(x) --> L - L = 0.

The special case is used in theorem conclusion as f(x) = (f(x) - g(x)) + g(x) --> 0 + L = L.

Hence the sandwich theorem is proved.

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