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You CAN Ace Calculus

17calculus > limits > pinching theorem

ATTENTION INSTRUCTORS: The new 2018 version of 17calculus will include changes to the practice problem numbering system. If you would like advance information to help you prepare for spring semester, send us an email at 2018info at 17calculus.com.

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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

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.

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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 \(\displaystyle{ \lim_{x \to 1}{f(x)} }\)


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

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


Practice A04

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



Level B - Intermediate

Practice B01



Practice B02



Practice B03



Practice B04



Practice B05



Practice B06



Practice B07



Practice B08



Practice B09



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