Sandwich Theorem

Limits are one of the most basic ideas in calculus, where one can learn how functions behave as they approach particular points. Of interest, though, is that some limits tend not to be as straightforward as finding the others, such that they could evaluate the functions. The application of the Sandwich theorem needs the real-valued functions to share the same domain of consideration.

In this article, we will cover the concept of the Sandwich Theorem. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of three questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2018, one in 2021, and one in 2023.

Sandwich Theorem

The Sandwich theorem is also known as the squeeze play theorem. It is typically used to find the limit of a function via comparison with two other functions whose limits are known or are easily calculated.

Let $f(x), g(x)$ and $h(x)$ be real functions such that $f(x) \leq g(x) \leq h(x)$ for all $x$ in neighbourhood of$\mathrm{x}=\mathrm{a}$. If $\lim\limits _{x \rightarrow a} f(x)=l=\lim\limits _{x \rightarrow a} h(x)$, then $\lim\limits _{x \rightarrow a} g(x)=l$.

Also, It also shares the same domain such that $f(x) \leq g(x) \leq h(x)$ for $\forall x$
- in the domain of definition states the following.
- For some real value of $a$, if $\lim\limits _{x \rightarrow a} f(x)=l=\lim\limits _{x \rightarrow a} h(x)$, then $\lim\limits _{x \rightarrow a} f(x)=l$
- wherein

Where $x \epsilon(a-\delta, a+\delta)$ and $\delta$ is very small.
The Floor Function indicates the greatest integer function $[G \mid F]$ denoted mathematically as $[P]$ for only real values. This function $[P]$ rounds downs the real number having any fractional or decimal part (if any) to the nearest integral value less than the indicated number.
Sandwich Theorem Proof:
Assume three real-valued functions $g(x), f(x)$, and $h(x)$ such that $g(x) \leq f(x) \leq h(x)$ and $\lim\limits _{x \rightarrow a} g(x)=\lim\limits _{x \rightarrow a} h(x)=L$.
Then by the definition of limits,

$
\begin{aligned}
& \lim\limits _{x \rightarrow a} g(x)=L \text { signifies } \forall \in>0, \exists \delta_1>0 \text { such that }|x-a|<\delta_1 \Rightarrow|g(x)-L|<\epsilon \\
& |x-a|<\delta_1 \Rightarrow-E<g(x)-L<\epsilon \ldots \text { (i) }
\end{aligned}
$

$\begin{aligned}
& \lim\limits _{x \rightarrow a} h(x)=L \text { signifies } \forall \in>0, \exists \delta_2>0 \text { such that }|x-a|<\delta_2 \Rightarrow|h(x)-L|<\epsilon \\
& |x-a|<\delta_2 \Rightarrow-E<h(x)-L<\epsilon \quad \ldots \text { (ii) }
\end{aligned}
$
Given, $g(x) \leq f(x) \leq h(x)$
Subtracting L from each side of the inequality

$
\begin{aligned}
& g(x)-L \leq f(x)-L \leq h(x)-L \\
& \text { Taking } \delta=\text { minimum }\left\{\delta_1, \delta_2\right\} \text {, Nowr }|x-a|<\delta, \\
& -\epsilon<g(x)-L \leq f(x)-L \leq h(x)-L<\epsilon \quad \text { [using (i) and (iii)] } \\
& -\in<f(x)-L<\epsilon \\
& \lim\limits _{x \rightarrow a} f(x)=L
\end{aligned}
$

Thus, this proved the Sandwich Theorem.

Solved Examples Based On Sandwich Theorem:

Example 1: $\lim\limits _{x \rightarrow \infty} \frac{[x]}{e^x}$ equals
1) 0
2) 1
3) 2
4) 3

Solution:
We know,
$\begin{aligned}
& x-1<[x] \leq x \Rightarrow \frac{x-1}{e^x}<\frac{[x]}{e^x} \leq \frac{x}{e^x} \\
& \lim\limits _{x \rightarrow \infty} \frac{x-1}{e^x}=\lim\limits _{x \rightarrow \infty} 1 / e^x=0 \\
& \lim\limits _{x \rightarrow \infty} \frac{x}{e^x}=\lim\limits _{x \rightarrow \infty} 1 / e^x=0 \quad \text { (using L hospital rule) } \\
& \therefore \lim\limits _{x \rightarrow \infty} \frac{x-1}{e^x}=\lim\limits _{x \rightarrow 0} \frac{x}{e^x}=0
\end{aligned}
$

$\therefore$ Using sandwich theorem

$\lim\limits _{x \rightarrow \infty} \frac{[x]}{e^x}$ =0

Hence, the answer is the option 1.
Example 2: $\lim\limits _{x \rightarrow 0} \frac{x^2}{e^{[x]}}$ equals:

1) 0

2) 1

3) 2

4) 3

Solution:

We know that

$
x-1<[x] \leq x
$
So, $e^{x-1}<e^{[x]} \leqslant e^x \Rightarrow \frac{x^2}{e^x} \leq \frac{x^2}{e^{[x]}}<\frac{x^2}{e^{x-1}}$
Now $\lim\limits _{x \rightarrow \infty} \frac{x^2}{e^{x-1}}=\lim\limits _{x \rightarrow \infty} \frac{2 x}{e^{x-1}}=\lim\limits _{x \rightarrow \infty} \frac{2}{e^{x-1}}=0$
and $\lim\limits _{x \rightarrow \infty} \frac{x^2}{e^x}=\lim\limits _{x \rightarrow \infty} \frac{2 x}{e^x}=\lim\limits _{x \rightarrow \infty} \frac{2}{e^x}=0$
(Both limits were calculated using L-Hospital's Rule )

$
\begin{aligned}
& \because \lim\limits _{x \rightarrow \infty} \frac{x^2}{e^{x-1}}=\lim\limits _{x \rightarrow \infty} \frac{x^2}{e^x}=0 \\
& \therefore \lim\limits _{x \rightarrow \infty} \frac{x^2}{e^{[x]}}=0
\end{aligned}
$

Hence, the answer is 0 .
Example 3: For each $t \in \mathrm{R}$, let $[\mathrm{t}]$ be the greatest integer less than or equal to $t$. Then $\left.\lim\limits _{x \rightarrow 0} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots .+\frac{15}{x}\right]\right)$.
[JEE MAIN 2018]

1) does not exist (in R).

2) is equal to 0.

3) is equal to 15

4) is equal to 120

Solution:

As we learned in

The SANDWICH THEOREM -

If $f(x) \leq g(x) \leq h(x)$ for every $(x)$ in the deleted neighbourhood of $(a)$

and $\lim\limits _{x \rightarrow a} f(x)=\lim\limits _{x \rightarrow a} h(x)=1$
Then $\lim\limits _{x \rightarrow a} g(x)$ is al so equal tol
$L=\lim\limits _{x \rightarrow 0+} x\left[\left(\frac{1}{x}\right)+\ldots+\left(\frac{15}{x}\right)\right]$
$L=\lim\limits _{x \rightarrow 0^{+}} x\left(\frac{1}{x}\right)+x\left(\frac{2}{x}\right)+\ldots+x\left(\frac{15}{x}\right)$
Take $\lim\limits _{x \rightarrow 0^{+}} x\left[\frac{1}{x}\right]$
$\operatorname{Let} \frac{1}{x}=y$
$+$
when $x \rightarrow 0, y \rightarrow \infty$
$L=\lim\limits _{y \rightarrow \infty} \frac{1}{y} y$
Let $y=u+p$, when $n \rightarrow \infty \quad, 0<p<1$

$
\lim\limits _{n \rightarrow \infty} \frac{n}{n+p}
$

$
\lim\limits _{\rightarrow \infty} \frac{1}{1+\frac{p}{n}}=1
$
Take $\lim\limits _{x \rightarrow 0^{+}} x\left[\frac{2}{x}\right]$
Let $\frac{2}{x}=y$
when $x \rightarrow 0, y \rightarrow \infty$

$
\begin{aligned}
& \lim\limits _{y \rightarrow \infty} \frac{2}{y} y=2 \\
& \lim\limits _{x \rightarrow 0+} x\left[\frac{1}{x}\right]+x\left[\frac{2}{x}\right]+\cdots+-+\left[\frac{15}{x}\right] \\
& =1+2+3+--++15 \\
& =\frac{n(n+1)}{2} \\
& =\frac{15 \times 16}{2} \\
& =120
\end{aligned}
$

Example 4: The value of $\lim\limits _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots+[n r]}{n^2}$where r is a non-zero real number and denoted the greatest integer less than or equal to r, is equal to : [JEE MAIN 2021]

1) $\frac{r}{2}$
2) $r$
3) $2 r$
4) 0

Solution:

As we know that

$
\begin{gathered}
\mathrm{r}-1<[\mathrm{r}] \leq \mathrm{r} \\
2 \mathrm{r}-1<[2 \mathrm{r}] \leq 2 \mathrm{r} \\
\vdots \\
\mathrm{nr}-1<[\mathrm{nr}] \leq \mathrm{nr}
\end{gathered}
$
Add all this

$
\begin{aligned}
& (\mathrm{r}+2 \mathrm{r}+\ldots \ldots+\mathrm{nr})-\mathrm{n}<[\mathrm{r}]+[2 \mathrm{r}]+\ldots \ldots+[\mathrm{nr}] \leq \mathrm{r}+2 \mathrm{r}+\ldots \ldots+\mathrm{nr}_{\text {Now }} \\
& \frac{(\mathrm{r}+2 \mathrm{r}+\ldots .+\mathrm{nr})-\mathrm{n}}{\mathrm{n}^2}<\frac{[\mathrm{r}]+[2 \mathrm{r}]+\ldots \ldots+[\mathrm{nr}]}{\mathrm{n}^2} \leq \frac{\mathrm{r}+2 \mathrm{r}+\ldots \ldots+\mathrm{nr}}{\mathrm{n}^2} \\
& \frac{\frac{n(n+1)}{2} \cdot r-n}{n^2}<\frac{[r]+[2 r]+\ldots .+[n r]}{n^2} \leq \frac{\frac{n(n+1)}{2} r}{n^2}
\end{aligned}
$
Now,
$\lim\limits _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2} \cdot r-n}{n^2}=\frac{r}{2}$
$\lim\limits _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}}{n^2}=\frac{r}{2}$
By sandwich theorem

$
\lim\limits _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^2}=\frac{r}{2}
$
Hence, the answer is the option 1.

Example $5: \lim\limits _{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{22 n+1}}\right)\right\}_{\text {is equal to: }}$
[JEE MAIN 2023]
1) $\frac{1}{\sqrt{2}}$
2) $\sqrt{2}$
3) 1
4) 0

Solution:
$
\begin{aligned}
& P=\lim\limits _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots \cdots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right) \\
& \text { Let } \\
& 2^{\frac{1}{2}}-2^{\frac{1}{3}} \quad \rightarrow \text { Smallest } \\
& 2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}} \rightarrow \text { Largest }
\end{aligned}
$
Using Sandwich theorem:

$
\begin{aligned}
& \left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n \leq P \leq\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n \\
& \binom{\operatorname{lie}^n b / w}{0 \text { and } 1}^n \\
& \lim\limits _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 \\
& \lim\limits _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n=0 \\
& \therefore \mathrm{P}=0
\end{aligned}
$

Hence, the answer is the option 4.

Summary

The sandwich theorem using limits is a useful part of calculus particularly used to find the limit of a function via comparison with two other functions whose limits are known or are easily calculated. Calculus was created to describe how the quantities change. The concept of limit is the cornerstone on which the development of calculus rests.