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You CAN Ace Calculus

17calculus > limits > pinching theorem

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

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

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

answer

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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