Question

what is the sandwich theorem in limits?

Answer 1

Suppose for all \(x \in (a,b)\):\[
g(x)\leq f(x)\leq h(x)
\]If for some \(c\in (a,b)\):\[
\lim_{x\to c}g(x) = L = \lim_{x\to c}h(x)
\]Then that means\[
\lim_{x\to c}f(x) = L
\]

Answer 2

In short:\[
\forall x \in (a,b)\quad g(x)\leq f(x)\leq h(x)\\
\exists c\in (a,b)\quad \lim_{x\to c}g(x) = L = \lim_{x\to c}h(x)
\implies
\lim_{x\to c}f(x) = L
\]

Answer 3

\(h(x)\) is the top bread and \(g(x)\) is the bottom bread.

Answer 4

And \(c\) is the position where you are trying to find the limit.

Answer 5

The sandwich theorem is when the limit from the left side approaching c is the same as the limit from the right side approaching c. Since f(x) is between the two limits, which are the same, f(x) must equal the value of the limit.

Answer 6

Here is an example:
\[Calculate \lim_{x \rightarrow 0} x ^{2}\cos(\frac{ 1 }{ x })\]
You cannot plug in x for obvious reasons so you use the sandwich theorem.

cos(x) lies between -1 and 1 therefore
\[-1 \le \cos(\frac{ 1 }{ x }) \le 1\]
because x can be any value including 1/x. Multiply by x^2.
\[-x ^{2} \le x ^{2} \cos(\frac{ 1 }{ x }) \le x ^{2} \]
Take the limit
\[\lim_{x \rightarrow 0}-x ^{2} \le \lim_{x \rightarrow 0}x ^{2} \cos(\frac{ 1 }{ x }) \le \lim_{x \rightarrow 0}x ^{2} \]
The left and right limits both equal 0 therefore the sandwich theorem states that the middle term must also equal 0.

Answer 7

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Answer 8

cool thanks for your answers!!!!

Answer 9

The squeeze theorem is a technical result that is very important in proofs in calculus and mathematical analysis. It is typically used to confirm the limit of a function via comparison with two other functions whose limits are known or easily computed. g(x) \leq f(x) \leq h(x) \,