The squeeze theorem allows us to find the limit of a function at a particular point, even when the function is not defined at that point. The way we do this is by showing that our function can be squeezed between two other functions at a given point and proving that the limits of these other functions are equal.

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## Step by step guide to determining limits using the squeeze theorem

**The squeeze theorem: **

We’ll assume our original function is \(h(x)\) and that it’s placed between two other functions, \(f(x)\) and and \(g(x)\), so:

\(\color{blue}{f(x)≤h(x)≤g(x)}\)

It’s also a given that when we approach the point of our interest, the limitations of our other two functions are equal, therefore this assumption is also made:

\(\color{blue}{lim_{x\to c}f(x)=lim_{x\to c}g(x)=L}\)

If we can show that both of the above statements are true, then we know that our original function has the same limit as the other two functions, and we say:

\(\color{blue}{lim_{x\to c}h(x)=L}\)

We don’t need to know what happens to \(h(x)\) at \(x=c\) since we don’t need to know. We only care about the limit, therefore all we need to know is what’s going on around \(x=c\).

### Determining Limits Using the Squeeze Theorem – Example 1:

Suppose there are three functions that \(f(x)≤ g(x) ≤ h(x)\) when \(x\) is near \(2\). Further, suppose \(f(x)=-\frac{1}{3}x^3+x^2-\frac{7}{3}\) and \(h(x)=cos(\frac{\pi}{2}x)\). Find \(lim_{x\to 2}g(x)\).

First, find \(f(x)\):

\(lim_{x\to 2}f(x)=lim_{x\to 2}-\frac{1}{3}x^3+x^2-\frac{7}{3}\)

\(=lim_{x\to 2}-\frac{1}{3}(2)^3+(2)^2-\frac{7}{3}\)

\(=-\frac{8}{3}+4-\frac{7}{3}\)

\(=\frac{-8+4(3)-7}{3}=\frac{-8+12-7}{3}\)

\(=-\frac{3}{3}=-1\)

Then, find \(h(x)\):

\(lim_{x\to 2}h(x)=lim_{x\to 2}cos(\frac{\pi}{2}x)\)

\(=lim_{x\to 2}cos(\frac{\pi}{2}2)\)

\(=cos{\pi}\)

\(=-1\)

Since \(f(x)≤ g(x) ≤ h(x)\) and \(lim_{x\to 2}f(x)=lim_{x\to 2}h(x)=-1\), the Squeeze Theorem guarantees \(lim_{x\to 2}g(x)=-1\).

## Exercises for Determining Limits Using the Squeeze Theorem

### Calculate the value of the following limit.

- \(\color{blue}{lim_{x\to 0} x^2sin\frac{1}{x}}\)
- \(\color{blue}{lim _{x\to \infty }\left(\frac{3x+cos^2\left(3x+1\right)}{7-4x}\right)}\)
- \(\color{blue}{lim _{x\to \infty }\left(\frac{3-cosx}{x+6}\right)}\)
- \(\color{blue}{lim _{x\to 2}\left(x^2+x-6\right)cos\left(\frac{1}{x-2}\right)}\)

- \(\color{blue}{0}\)
- \(\color{blue}{-\frac{3}{4}}\)
- \(\color{blue}{0}\)
- \(\color{blue}{0}\)

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