## Squeeze Theorem

### Topics You Need To Understand For This Page

 precalculus basics of limits finite limits

### Page Tools and Related Topics

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### Video List

A02  squeeze thm 2

B02  squeeze thm 3

B03  squeeze thm 4

B04  squeeze thm 5

B05  squeeze thm 6

A03  squeeze thm 1

B06  squeeze thm 2

B07  squeeze thm 3

B08  squeeze thm 4

B09  squeeze thm 5

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Alternate Names For The Squeeze Theorem

The Sandwich Theorem

The Pinching Theorem

Squeeze 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 Squeeze Theorem is a powerful theorem that allows us to prove several important limits, including this trig limit.

 $$\displaystyle{ \lim_{\theta \to 0}{\frac{sin(\theta)}{\theta}} = 1 }$$ proof

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. 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.

This video explains the squeeze theorem with a graph and it is a good place to start.

Here is a great video showing the proof of $$\displaystyle{ \lim_{x \to 0}{\left[ xsin(1/x)\right]} = 0 }$$ using the squeeze theorem (which he calls the pinching theorem). He does some unusual things here, so it is important to watch this video to see this technique.

Squeeze Theorem At Infinity

In the above discussion, we used the squeeze 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.

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### Squeeze Theorem Resources

Practice Problems

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

 Level A - Basic

Practice A01

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

Practice A02

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

Practice A03

If   $$4x-9 \leq f(x) \leq x^2-4x+7$$
for all $$x$$, evaluate $$\displaystyle{ \lim_{x \to 4}{f(x)} }$$

##### Practice Problem [ A01 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 1}{f(x)} = 1 }$$

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##### Practice Problem [ A02 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 1}{f(x)} = 4 }$$

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##### Practice Problem [ A03 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 4}{f(x)} = 7 }$$

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 Level B - Intermediate

Practice B01

$$\displaystyle{ \lim_{x \to 0}{ \left[ x^2 \cdot \cos(1/x^2 ) \right] } }$$

Practice B02

$$\displaystyle{ \lim_{x \to 0}{ x^4 sin(3/x) } }$$

Practice B03

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

Practice B04

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

##### Practice Problem [ B01 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ \left[ x^2 \cdot \cos(1/x^2 ) \right] } = 0 }$$

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##### Practice Problem [ B02 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ x^4 sin(3/x) } = 0 }$$

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##### Practice Problem [ B03 ] Solution

Final Answer
$$\displaystyle{ \lim_{n \to \infty}{ \frac{1}{n^3} \sin(n^2) } = 0 }$$

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##### Practice Problem [ B04 ] Solution

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

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Practice B05

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

Practice B06

$$\displaystyle{ \lim_{x \to 0}{ x^2 \cos(10x) } }$$

Practice B07

$$\displaystyle{ \lim_{x \to 0}{ x^2 cos (1/\sqrt[3]{x}) } }$$

Practice B08

$$\displaystyle{ \lim_{x \to 0}{ \sqrt[3]{x} sin(1/x) } }$$

##### Practice Problem [ B05 ] Solution

Final Answer
$$\displaystyle{ \lim_{n \to \infty}{ n^{-n^3} } = 0 }$$

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##### Practice Problem [ B06 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ x^2 \cos(10x) } = 0 }$$

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##### Practice Problem [ B07 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ x^2 cos (1/\sqrt[3]{x}) } = 0 }$$

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##### Practice Problem [ B08 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ \sqrt[3]{x} sin(1/x) } = 0 }$$

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Practice B09

$$\displaystyle{ \lim_{x \to 0}{ x^2 sin(1/x^2) } }$$

##### Practice Problem [ B09 ] Solution

Final Answer
$$\displaystyle{ \lim_{x \to 0}{ x^2 sin(1/x^2) } = 0 }$$

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Squeeze Theorem Proofs

### Proof of sin(x)/x Limit

prove that

$$\displaystyle{ \lim_{\theta \to 0}{\frac{sin(\theta)}{\theta}} = 1 }$$

An informal proof is shown in this video.

 video by Khan Academy