\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Limits - Pinching Theorem

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The Pinching Theorem is also referred to as The Sandwich Theorem or The Squeeze Theorem.

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.

PatrickJMT - The Squeeze Theorem for Limits, Example 1 [7min-12secs]

video by PatrickJMT

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.

Dr Chris Tisdell - Limit of a function: Pinching theorem with streamlined method of solution [4min-22secs]

video by Dr Chris Tisdell

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

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

470 video

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

472 video

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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

477 video

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\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)

Problem Statement

Evaluate the limit using the pinching theorem. Give your answer in exact form. \(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)

Final Answer

\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}=0}\)

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

1948 video

Final Answer

\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}=0}\)

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Intermediate

\(\displaystyle{\lim_{x\to0}{\left[x^2\cdot\cos(1/x^2)\right]}}\)

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

471 video

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

473 video

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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

474 video

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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

475 video

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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

476 video

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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

478 video

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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

479 video

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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

480 video

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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

481 video

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pinching theorem 17calculus youtube playlist

Here is a playlist of the videos on this page.

You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

Wikipedia - Squeeze Theorem

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Pinching Theorem

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Practice Instructions

Unless otherwise instructed, evaluate these limits using the pinching theorem. Give your answers in exact form.

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