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17Calculus Limits - Pinching Theorem

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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

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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

close solution

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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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

1948 video

video by Dr Chris Tisdell

Final Answer

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

close solution

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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

close solution

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

Problem Statement

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

Solution

473 video

video by PatrickJMT

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

Problem Statement

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

Solution

474 video

video by PatrickJMT

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

Problem Statement

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

Solution

475 video

video by PatrickJMT

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\(\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}\)

Problem Statement

\(\displaystyle{\lim_{n\to\infty}{n^{-n^3}}}\)

Solution

476 video

video by PatrickJMT

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

Problem Statement

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

Solution

478 video

video by Krista King Math

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

Problem Statement

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

Solution

479 video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

480 video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

481 video

video by Krista King Math

close solution

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