The Pinching Theorem is also referred to as The Sandwich Theorem or The Squeeze Theorem.
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Pinching Theorem
For an interval I containing a point a, if we have three functions, \(f(x), g(x), h(x)\) all three defined on the interval I, except at a and we have \(\displaystyle{ g(x) \leq f(x) \leq h(x) }\) for all points in I except possibly at \(x=a\) and we know that \(\displaystyle{ \lim_{x \to a}{g(x)} = \lim_{x \to a}{h(x)} = L }\) Then \(\displaystyle{ \lim_{x \to a}{f(x)} = L }\) |
The Pinching Theorem is a powerful theorem that allows us to determine several important limits, including this important trig limit.
\(\displaystyle{ \lim_{\theta \to 0}{\frac{\sin(\theta)}{\theta}} = 1 }\) |
To help you get your head around this theorem, here is a graph that intuitively shows you the idea of the proof of the above limit. (See this separate page for two additional proofs of this limit.) The red line is the function
\( \color{#BF003F}{ f(x) = \sin(x)/x }\)
, the green line is \( \color{#007F7F}{g(x) = \cos(x) }\) and the blue line is \(\color{#0000FF}{h(x) = 1}\). Notice that near, i.e. within an interval, around \(x=0\),
\( \color{#007F7F}{g(x)} \leq \color{#BF003F}{f(x)} \leq \color{#0000FF}{h(x)}\) and
\(\displaystyle{ \color{#007F7F}{\lim_{x \to 0}{g(x)} = 1} }\) |
\(\displaystyle{ \color{#0000FF}{\lim_{x \to 0}{h(x)} = 1} }\) |
which means \(\displaystyle{ \color{#BF003F}{\lim_{x \to 0}{f(x)} = 1} }\)
The trick comes in when you have to find two functions \(g(x)\) and \(h(x)\) that satisfy the theorem. However, if you graph \(f(x)\), sometimes you will be able to find a couple of functions that work. However, there is one idea that will sometimes work. If you have the trig functions \(\sin(x)\) or \(\cos(x)\), these functions are always between \(-1\) and \(+1\). So setting up an inequality with those will sometimes give you a place to start.
Okay, let's look at some videos to explain this in more detail. The first video explains the pinching(squeeze) theorem with a graph and it is a good place to start.
video by PatrickJMT |
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Here is a great video showing the proof of \(\displaystyle{ \lim_{x \to 0}{\left[ x\sin(1/x)\right]} = 0 }\) using the pinching theorem. He does some unusual things here, so it is important to watch this video to see this technique.
video by Dr Chris Tisdell |
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Pinching Theorem At Infinity
In the above discussion, we used the pinching theorem when the limit variable was going to a finite number. However, we can apply the same technique when the limit variable is going to infinity, i.e. we have the limit \(\displaystyle{ \lim_{x \to \infty}{f(x)} }\). We just need to show that the function is bounded above and below by two functions and that the limits of both bounding functions goes to the same value as x goes to infinity. You will find some practice problems below demonstrating this technique.
Practice
Unless otherwise instructed, evaluate these limits using the pinching theorem. Give your answers in exact form.
Basic |
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If \(3x \leq f(x) \leq x^3+2\) on \([0,2]\), evaluate \(\displaystyle{ \lim_{x \to 1}{f(x)} }\) using the pinching theorem.
Problem Statement
If \(3x \leq f(x) \leq x^3+2\) on \([0,2]\), evaluate \(\displaystyle{ \lim_{x \to 1}{f(x)} }\) using the pinching theorem. Give your answer in exact form.
Solution
video by PatrickJMT |
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If \(4\leq f(x)\leq x^2+6x-3\) for all \(x\), evaluate \(\displaystyle{ \lim_{x \to 1}{f(x)} }\) using the pinching theorem.
Problem Statement
If \(4\leq f(x)\leq x^2+6x-3\) for all \(x\), evaluate \(\displaystyle{ \lim_{x \to 1}{f(x)} }\) using the pinching theorem. Give your answer in exact form.
Solution
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Evaluate \(\displaystyle{\lim_{x\to4}{f(x)}}\) if, for all \(x\), \(4x-9\leq f(x)\leq x^2-4x+7\) using the pinching theorem.
Problem Statement
Evaluate \(\displaystyle{\lim_{x\to4}{f(x)}}\) if, for all \(x\), \(4x-9\leq f(x)\leq x^2-4x+7\) using the pinching theorem. Give your answer in exact form.
Solution
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\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)
Problem Statement |
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Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)
Final Answer |
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\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}=0}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)
Solution
Final Answer
\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}=0}\)
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Intermediate |
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\(\displaystyle{\lim_{x\to0}{\left[x^2\cdot\cos(1/x^2)\right]}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{\left[x^2\cdot\cos(1/x^2)\right]}}\)
Solution
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to0}{x^4\sin(3/x)}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{x^4\sin(3/x)}}\)
Solution
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\(\displaystyle{\lim_{n\to\infty}{\frac{1}{n^3}\sin(n^2)}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\frac{1}{n^3}\sin(n^2)}}\)
Solution
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\(\displaystyle{\lim_{n\to\infty}{\frac{(-1)^n+n^2}{n^2}}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\frac{(-1)^n+n^2}{n^2}}}\)
Solution
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\(\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}\)
Solution
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\(\displaystyle{\lim_{x\to0}{x^2\cos(10x)}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{x^2\cos(10x)}}\)
Solution
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\(\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt[3]{x})}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt[3]{x})}}\)
Solution
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\(\displaystyle{\lim_{x\to0}{\sqrt[3]{x}\sin(1/x)}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{\sqrt[3]{x}\sin(1/x)}}\)
Solution
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\(\displaystyle{\lim_{x\to0}{x^2\sin(1/x^2)}}\)
Problem Statement
Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{\lim_{x\to0}{x^2\sin(1/x^2)}}\)
Solution
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