## Links forward

### The pinching theorem

One very useful argument used to find limits is called the **pinching theorem**. It essentially says that if we can `pinch' our limit between two other limits which have a common value, then this common value is the value of our limit.

Thus, if we have

\[ g(x) \leq f(x) \leq h(x), \quad \text{for all } x, \]and \(\displaystyle \lim_{x\to a} g(x) = \lim_{x\to a} h(x)= L\), then \(\displaystyle \lim_{x\to a} f(x) = L\).

Here is a simple example of this.

To find \(\displaystyle \lim_{n\to \infty} \dfrac{n!}{n^n}\), we can write

\begin{align*} \dfrac{n!}{n^n} &= \dfrac{n}{n} \times \dfrac{n-1}{n} \times \dfrac{n-2}{n} \times \dots \times \dfrac{3}{n} \times \dfrac{2}{n} \times \dfrac{1}{n}\\ &\leq 1\times 1\times 1\times \dots \times 1 \times 1 \times \dfrac{1}{n} = \dfrac{1}{n}, \end{align*} where we replaced every fraction by 1 except the last. Thus we have \(0 \leq \dfrac{n!}{n^n} \leq \dfrac{1}{n}\). Since \(\displaystyle \lim_{n\to \infty} \dfrac{1}{n} = 0\), we can conclude using the pinching theorem that \(\displaystyle \lim_{n\to \infty} \dfrac{n!}{n^n}=0\).Other examples will be found in later modules. In particular, the very important limit

\[ \dfrac{\sin x}{x} \to 1 \quad \text{as} \quad x \to 0 \](where \(x\) is expressed in radians) will be proven using the pinching theorem in the module The calculus of trigonometric functions.