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You CAN Ace Calculus

17calculus > limits > pinching theorem

### Calculus Main Topics

Limits

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

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.

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

### 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.

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. 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

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

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.

### Search 17Calculus

Practice Problems

Instructions - Unless otherwise stated, evaluate the following limits using the pinching 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 $$\displaystyle{ \lim_{x \to 1}{f(x)} }$$

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

 Level B - Intermediate

Practice B01

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

solution

Practice B02

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

solution

Practice B03

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

solution

Practice B04

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

solution

Practice B05

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

solution

Practice B06

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

solution

Practice B07

$$\displaystyle{\lim_{x\to0}{x^2\cos(1/\sqrt[3]{x})}}$$

solution

Practice B08

$$\displaystyle{\lim_{x\to0}{\sqrt[3]{x}\sin(1/x)}}$$

solution

Practice B09

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

solution

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