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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
prove that 

\(\displaystyle{ \lim_{\theta \to 0}{\frac{\sin(\theta)}{\theta}} = 1 }\) 
An informal proof is shown in this video.
video by Khan Academy 
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 \(4\leq f(x)\leq x^2+6x3\) 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, \(4x9\leq f(x)\leq x^24x+7\)  
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 