## 17Calculus Limits - Pinching Theorem

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

Instructions - Unless otherwise stated, evaluate the following limits using the pinching theorem. Give your answers in exact form.

Basic Problems

If $$3x \leq f(x) \leq x^3+2$$ on $$[0,2]$$, evaluate $$\displaystyle{ \lim_{x \to 1}{f(x)} }$$.

Problem Statement

If $$3x \leq f(x) \leq x^3+2$$ on $$[0,2]$$, evaluate $$\displaystyle{ \lim_{x \to 1}{f(x)} }$$.

Solution

### 470 video

video by PatrickJMT

If $$4\leq f(x)\leq x^2+6x-3$$ for all $$x$$, evaluate $$\displaystyle{ \lim_{x \to 1}{f(x)} }$$.

Problem Statement

If $$4\leq f(x)\leq x^2+6x-3$$ for all $$x$$, evaluate $$\displaystyle{ \lim_{x \to 1}{f(x)} }$$.

Solution

### 472 video

video by PatrickJMT

Evaluate $$\displaystyle{\lim_{x\to4}{f(x)}}$$ if, for all x, $$4x-9\leq f(x)\leq x^2-4x+7$$.

Problem Statement

Evaluate $$\displaystyle{\lim_{x\to4}{f(x)}}$$ if, for all x, $$4x-9\leq f(x)\leq x^2-4x+7$$.

Solution

### 477 video

video by Krista King Math

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

Problem Statement

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

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

Problem Statement

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

Solution

### 1948 video

video by Dr Chris Tisdell

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

Intermediate Problems

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

Problem Statement

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

Solution

### 471 video

video by PatrickJMT

$$\displaystyle{\lim_{x\to0}{x^4\sin(3/x)}}$$

Problem Statement

$$\displaystyle{\lim_{x\to0}{x^4\sin(3/x)}}$$

Solution

### 473 video

video by PatrickJMT

$$\displaystyle{\lim_{n\to\infty}{\frac{1}{n^3}\sin(n^2)}}$$

Problem Statement

$$\displaystyle{\lim_{n\to\infty}{\frac{1}{n^3}\sin(n^2)}}$$

Solution

### 474 video

video by PatrickJMT

$$\displaystyle{\lim_{n\to\infty}{\frac{(-1)^n+n^2}{n^2}}}$$

Problem Statement

$$\displaystyle{\lim_{n\to\infty}{\frac{(-1)^n+n^2}{n^2}}}$$

Solution

### 475 video

video by PatrickJMT

$$\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}$$

Problem Statement

$$\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}$$

Solution

### 476 video

video by PatrickJMT

$$\displaystyle{\lim_{x\to0}{x^2\cos(10x)}}$$

Problem Statement

$$\displaystyle{\lim_{x\to0}{x^2\cos(10x)}}$$

Solution

### 478 video

video by Krista King Math

$$\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt{x})}}$$

Problem Statement

$$\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt{x})}}$$

Solution

### 479 video

video by Krista King Math

$$\displaystyle{\lim_{x\to0}{\sqrt{x}\sin(1/x)}}$$

Problem Statement

$$\displaystyle{\lim_{x\to0}{\sqrt{x}\sin(1/x)}}$$

Solution

### 480 video

video by Krista King Math

$$\displaystyle{\lim_{x\to0}{x^2\sin(1/x^2)}}$$

Problem Statement

$$\displaystyle{\lim_{x\to0}{x^2\sin(1/x^2)}}$$

Solution

### 481 video

video by Krista King Math

### pinching theorem 17calculus youtube playlist

You CAN Ace Calculus

 precalculus basics of limits finite limits

Wikipedia - Squeeze Theorem

### Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

### 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. 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.
 Pinching Theorem Pinching Theorem At Infinity Practice

Join Amazon Student - FREE Two-Day Shipping for College Students Save Up To 50% Off SwissGear Backpacks Plus Free Shipping Over \$49 at eBags.com! Prime Student 6-month Trial When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.