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Squeeze Theorem, also known as Sandwich Theorem, is a theorem used to find the limits of a function that is squeezed between two functions. The modern Squeeze form was given by Carl Friedrich Gauss. In this article, we will discuss the Squeeze theorem and the steps to apply and prove the Squeeze theorem in questions. Squeeze theorem proof and examples will be elaborated in this article in simpler form for better clarity of the topic and how to do Squeeze theorem in function will be discussed here.
Contents
History of Carl Friedrich Gauss
Carl Friedrich Gauss
Name: Carl Friedrich Gauss
Born: 30 April 1777
Died: 23 February 1855
Field: Mathematics
Nationality: German
Statement of Squeeze Theorem
According to Squeeze Theorem, if g(x), f(x), h(x) are three functions such that [g(x) leq f(x) leq h(x)] and at any point a [mathop {{rm{;}}lim }limits_{x to a} g(x) = mathop {{rm{;}}lim }limits_{x to a} h(x) = L ], then [mathop {{rm{;}}lim }limits_{x to a} f(x) = L].
Squeeze Theorem Proof
Assume that $g(x) leq f(x) leq h(x)$ and $mathop {{rm{;}}lim }limits_{x to a} g(x)=mathop {{rm{;}}lim }limits_{x to a} h(x)=L$.
Squeeze Theorem
Using the definition of limits,
$Rightarrow mathop {{rm{;}}lim }limits_{x to a} g(x)=L text { means that } forall in>0, exists delta_{1}>0$ such that
$Rightarrow|x-a|<delta_{1}$
$Rightarrow |x-a| < delta_{1}$
$Rightarrow|g(x)-L|< epsilon$
$Rightarrow – epsilon < g(x)-L< epsilon ldots text { (1) } $(expanding using property of modulus)
$Rightarrow lim _{x rightarrow a} h(x)=L text { using definition of limits again } forall in>0, exists delta_{2}>0 text { such that }$
$Rightarrow |x-a| < delta_{2} $
$Rightarrow |x-a| < delta_{2} $
$Rightarrow |h(x)-L| < epsilon$
$Rightarrow – epsilon < h(x)-L < epsilon ldots text { (2) }$ (expanding using property of modulus)
Now, it is given that $g(x) leq f(x) leq h(x)$.
Subtracting $L$ from each side of the above expression, we get
$Rightarrow g(x)-L leq f(x)-L leq h(x)-L$
Let us choose $delta=minimum left{delta_{1}, delta_{2}right}$,
we have, whenever $|x-a| < delta$,
$Rightarrow – epsilon < g(x)-L leq f(x)-L leq h(x)-L <epsilon text { (using (1) and (2)) }$
So, from the above expression, we have
$Rightarrow – epsilon < f(x)-L < epsilon$
$Rightarrow mathop {{rm{;}}lim }limits_{x to a} f(x) = L$
Hence, the squeeze theorem is proved.
Limitations of Squeeze Theorem
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The squeeze theorem is not applicable if left and right function limits are not equal.
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The squeeze Theorem does not apply if we know the limits of any of two functions other than both extreme functions.
Applications of Squeeze Theorem
Squeeze Theorem Examples
1. Evaluate the limit $mathop {{rm{;}}lim }limits_{x to 0}left(x cdot cos left(dfrac{1}{x}right)right)$, if it exists.
Ans: We know that $-1 leq cos left(frac{1}{x}right) leq 1$ for all $x neq 0$.
Then, $-x leq x cdot cos left(frac{1}{x}right) leq x$, so
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}(-x) leq mathop {{rm{;}}lim }limits_{x to 0} left(x cdot cos left(frac{1}{x}right)right) leq mathop {{rm{;}}lim }limits_{x to 0} x$
Since $mathop {{rm{;}}lim }limits_{x to 0}(-x)=0=mathop {{rm{;}}lim }limits_{x to 0} x$,
we see that
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}left(x cdot cos left(frac{1}{x}right)right)=0$.
2. Evaluate the limit using the Squeeze Theorem:
$mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos frac{5}{x}$
Ans: We know that $-1 leq cos frac{1}{x} leq 1$.
Next, multiplying the inequality by $x^{2}$, we have
$Rightarrow-x^{2} leq x^{2} cos frac{5}{x} leq x^{2}$
Take the limit of each part of the inequality.
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}left(-x^{2}right) leq mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos frac{5}{x} leq mathop {{rm{;}}lim }limits_{x to 0} x^{2}$
Next, we know that
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}left(-x^{2}right)=0 text { and } mathop {{rm{;}}lim }limits_{x to 0}left(x^{2}right)=0$.
Thus, we have
$Rightarrow 0 leq mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos frac{5}{x} leq 0$
So, $mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos frac{5}{x}=0$
3. Evaluate the limit using the Squeeze Theorem:
$mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos (10 x)$
Ans: We know that $-1 leq cos (10 x) leq 1$.
Next, multiplying the inequality by $x^{2}$, we have
$Rightarrow-x^{2} leq x^{2} cos (10 x) leq x^{2}$
Take the limit of each part of the inequality.
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}left(-x^{2}right) leq mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos (10 x) leq mathop {{rm{;}}lim }limits_{x to 0} x^{2}$
Next, we know that
$Rightarrow mathop {{rm{;}}lim }limits_{x to 0}left(-x^{2}right)=0 text { and } mathop {{rm{;}}lim }limits_{x to 0}left(x^{2}right)=0$.
Thus, we have
$Rightarrow 0 leq mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos (10 x) leq 0$
So, $mathop {{rm{;}}lim }limits_{x to 0} x^{2} cos (10 x)=0$
Conclusion
In the article, we have discussed the detailed proof of the Squeeze Theorem and its proof. Squeeze Theorem in simple terms says that if you are stuck between two things, then you will go the same way the two things are going. In all, we can say that Squeeze Theorem is a fantastic theorem in calculus and we can say that calculus needs limits and limits needs Squeeze Theorem.
Important Formulas to Remember
If $g(x), f(x), h(x)$ are such that $g(x) leq f(x) leq h(x)$, and $mathop {{rm{;}}lim }limits_{x to a} g(x)=mathop {{rm{;}}lim }limits_{x to a} h(x)=L$ then $mathop {{rm{;}}lim }limits_{x to a} f(x)=L$
Important Points to Remember
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