Sandwich Theorem

The Sandwich Theorem is a theorem that is used to solve limits 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 Big-Theta time complexity of an algorithm.

Definition §

Sandwich Theorem

Let $f(x),g(x),h(x)$ all be real-valued function that operate over a domain $D$. Then if we have

• $g(x)\le f(x)\le h(x)$ for all $x\in D$. That is, $f(x)$ is bounded between $g(x)$ and $h(x)$
• ${\lim }_{x\to a}g(x)=L={\lim }_{x\to a}h(x)$

Then:

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

In other words, if $f(x)$ is strictly bound between $g(x)$ and $h(x)$ over a domain and both limits 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 $h(x)$.

In Two Dimensions §

In two dimensions i.e. for bivariate functions, the sandwich theorem still holds, albeit slightly modified:

• $g(x,y)\le f(x,y)\le h(x,y)\forall (x,y)\in D\subseteq {\mathbb{R}}^2$
• ${\lim }_{(x,y)\to (a,b)}g(x)={\lim }_{(x,y)\to (a,b)}h(x)=L$

Then we have:

$$\underset{(x,y)\to (a,b)}{\lim }f(x,y)=L$$

\usepackage{pgfplots}
\pgfplotsset{width=15cm,compat=1.16}
 \begin{document}
 
 \begin{tikzpicture}
 \begin{axis}[colormap/bluered]
 \addplot3[
 	mesh,
 	samples=20,
 	domain=-2:2
 ]
 {0};
 \addplot3[
 	mesh,
 	samples=20,
 	domain=-2:2,
 ]
 {x^2 + y^2};
 \addplot3[
 	mesh,
 	samples=20,
 	domain=-2:2,
 ]
 {-x^2-y^2};

 \end{axis}
 \end{tikzpicture}
 
 \end{document}

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

Evaluate the limit:

$$\underset{x\to 0}{\lim }f(x)f(x)=x\sin (\frac{1}{x})$$

Solution domain of the sine function is $[-1,1]$. Hence:

The

$$-1\le \sin (\frac{1}{x})\le 1$$

We can use this information to determine an upper and lower bound:

$$\begin{array}{rl}-1& \le \sin (\frac{1}{x})\le 1\\ -|x|& \le |x|\sin (\frac{1}{x})\le |x|\end{array}$$

Since |x| and -|x| are defined to be upper and lower bound functions, and their Limit (Calculus) 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, ${\lim }_{x\to 0}f(x)=0$

• Another important function to use for bounds is the absolute value function, $|x|$.

2: More complex functions

Evaluate the limit:

$$\underset{x\to \infty }{\lim }f(x)f(x)=\sqrt{x^2+1}-x$$

Solution $f(x)=\sqrt{x^2+1}-x\times \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$ (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.

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).

Bivariate §

3: Bivariate Sandwich Theorem ^e3

Evaluate

$$\underset{(x,y)\to (0,0)}{\lim }(f(x))\text{ where }f(x)=\frac{x^2}{\sqrt{x^2+y^2}}$$

Solution $g(x,y)$ and an upper bound $h(x,y)$. First note that the domain of $f$ is ${\mathbb{R}}^2\setminus \{0,0\}$, because at $(0,0)$, we would have the denominator being zero, and we cannot divide by zero. Hence the domain $D={\mathbb{R}}^2\setminus \{0,0\}$

We need to find a lower bound

• Let $g(x,y)=0$. Since $f(x)$ is undefined for $(0,0)$, and has sums of squared terms, we know it is also $>>0$. Hence $g(x,y)\le f(x,y)$
• We can take $h(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2}}$. This is essentially the adding $\frac{y^2}{\sqrt{x^2+y^2}}$ to $f(x)$. So by ordered field axioms, we have $h(x,y)\ge f(x,y)$
• Thus we have $g(x,y)\le f(x,y)\le h(x,y)$

We can evaluate the limits of $g(x)$ and $h(x)$:

$$\underset{(x,y)\to (0,0)}{\lim }(g(x))=\underset{(x,y)\to (0,0)}{\lim }(0)=0$$

$$\underset{(x,y)\to (0,0)}{\lim }(h(x))=\underset{(x,y)\to (0,0)}{\lim }(\frac{x^2+y^2}{\sqrt{x^2+y^2}})=\underset{(x,y)\to (0,0)}{\lim }(\sqrt{x^2+y^2})=0$$

Thus, the limits of the bounds exist and are equal. Thus, by the Sandwich theorem, we have:

$$\underset{(x,y)\to (0,0)}{\lim }(f(x,y))=0$$

Visuals

Resources §

• Squeeze Theorem - Organic Chemistry Tutor