Sandwich Theorem

The Sandwich Theorem is a theorem that is used to solve Limit that would otherwise be considered undefined. It works on the principle that if a function is ‘sandwiched’ between two other functions for a period, the value it converges (i.e. the limit) will also between the Limit of the two functions.

It also uses a similar principle to finding the average big-O time of an algorithm.

Definition §

Let $f(x),g(x),h(x)$ all be real-valued functions that operate over a domain of x, where $x\ne a$:

$$Ifg(x)\le f(x)\le h(x)$$

$$and\underset{x\to a}{\lim }g(x)=L=\underset{x\to a}{\lim }h(x)$$

$$then\underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }h(x)=\underset{x\to a}{\lim }g(x)=L$$

In other words, if f(x) is strictly bound between g(x) and h(x) over a domain and both Limit of g(x) and h(x) exist and are equal, then the limit of f(x) is equal to the Limit of g(x) and h(x).

Examples §

The trig functions sin and cos are a godsend here, as they are always bound between [-1,1] so any function multiplied by them can be used as g(x) and h(x):

1: Trig Functions

Let $f(x)=x\sin (\frac{1}{x})$. Evaluate ${\lim }_{x\to 0}f(x)$

Since $\sin (\frac{1}{x})$ is always bounded between -1 and 1, we can use it’s domain to determine an upper and lower bound function: $sin(\frac{1}{x})\in [1,-1]$ $⟹|x|\sin (\frac{1}{x})\in [-x,x]$ $⟹-|x|\le f(x)\le |x||x\ne 0$ Since |x| and -|x| are defined to be upper and lower bound functions, and their Limit exist and are equal at 0, trivially, then using the Sandwich Theorem:

${\lim }_{x\to 0}f(x)={\lim }_{x\to 0}|x|={\lim }_{x\to 0}-|x|=L=0$ Hence, ${\lim }_{x\to 0}f(x)=0$

2: More complex functions

Let $f(x)=\sqrt{x^2+1}-x$. Evaluate ${\lim }_{x\to \infty }f(x)$
At first, this limit might look daunting without the presence of any trig functions. However, it can still be simplified (or at least converted into a form which allows the use of the sandwich theorem). $f(x)=\sqrt{x^2+1}-x\times \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$ (Multiplying by the conjugate) $=\frac{x^2+1-x^2}{\sqrt{x^2+1}-x}$ $=\frac{1}{\sqrt{x^2+1}-x}$
Now, to use the sandwich theorem, we need to find 2 appropriately bounded functions to act as upper and lower bound. Let’s call the lower bound $g(x)$ and the upper bound $h(x)$.
$\frac{1}{\sqrt{x^2+1}-x}\ge \frac{1}{\sqrt{x^2+1}}\ge \frac{1}{\sqrt{x^2+x^2}}=\frac{1}{\sqrt{2}|x|}$ So, our lower bound function $g(x)=\frac{1}{\sqrt{2}|x|}$ $\frac{1}{\sqrt{x^2+1}-x}\le \frac{1}{\sqrt{x^2+1}-2\sqrt{x^2}}\le \frac{1}{\sqrt{x^2}-2|x|}\le \frac{1}{|x|-2|x|}=\frac{-1}{|x|}$ So, our upper bound function $h(x)=\frac{-1}{|x|}$
Now, to use the sandwich theorem: ${\lim }_{x\to \infty }\frac{1}{\sqrt{2}\times |x|}=\frac{1}{\sqrt{2}}{\lim }_{x\to \infty }\left(\frac{1}{|x|}\right)=0$ (By Standard Limits, as $\frac{1}{|x|}=\frac{1}{x}\forall x>0$) Similarly, ${\lim }_{x\to \infty }-\frac{1}{|x|}=-{\lim }_{x\to \infty }\frac{1}{|x|}=0$
Hence, as ${\lim }_{x\to \infty }(g(x))={\lim }_{x\to \infty }(h(x))$ and $g(x)\le f(x)\le h(x)$, ${\lim }_{x\to \infty }(f(x))=0$ via Sandwich Theorem.

Resources §