17Calculus Limits - Pinching Theorem
The Pinching Theorem is also referred to as The Sandwich Theorem or The Squeeze Theorem.
Recommended Books on Amazon (affiliate links) | ||
|---|---|---|
  |
  |
  |
Join Amazon Student - FREE Two-Day Shipping for College Students | ||
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 |
|---|
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}}\)
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 |
|---|
\(\displaystyle{ \lim_{n\to\infty}{\frac{\cos^2n}{n}}=0}\)
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
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 |
|---|
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{x\to0}{x^4\sin(3/x)}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{n\to\infty}{\frac{1}{n^3}\sin(n^2)}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{n\to\infty}{\frac{(-1)^n+n^2}{n^2}}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{x\to0}{x^2\cos(10x)}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt[3]{x})}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{x\to0}{\sqrt[3]{x}\sin(1/x)}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
\(\displaystyle{\lim_{x\to0}{x^2\sin(1/x^2)}}\)
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
|
close solution
|
Log in to rate this practice problem and to see it's current rating. |
|---|
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 |
|---|
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
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
To bookmark this page and practice problems, log in to your account or set up a free account.
Search Practice Problems
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
|
| |
free ideas to save on books |
|---|
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. |
Page Sections
How to Develop a Brilliant Memory Week by Week: 50 Proven Ways to Enhance Your Memory Skills |
|---|
Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over $49 at eBags.com! |
Practice Instructions
Unless otherwise instructed, evaluate these limits using the pinching theorem. Give your answers in exact form.
