# Sandwich Theorem of Limits

This is an important concept for Class 11 NCERT Math and is used to solve different problems related to Limits and Derivative. Sandwich Theorem is also called as Squeeze Theorem or Pinch Theorem

## Concept of Sandwich Theorem

Consider there are two functions f(x) and g(x) such that the function f(x) is always greater than equal to g(x)
i.e f(x) \geqq g(x)

Observe given graphical representation

In the above illustration you can see two functions f(x) and g(x)
Note that the value of function f(x) is always greater than the function g(x), except at point P where both function have approximately equal value.

Now let us consider third function h(x) which lies between f(x) and g(x) such that:
g( x) \leqq h( x) \leqq f( x)
Hence the function h(x) is greater than equal to g(x) and less than equal to f(x)

Now observe the illustration with the three functions

Clearly the function h(x) lies between f(x) and g(x)
But at point P, all the function meet and have equal limit value.
i.e. \lim f(x) =\lim g( x) =lim\ h(x)

Now the Sandwich/Squeeze theorem states that:
When there are three functions f(x), g(x) and h(x) such that
g( x) \leqq h( x) \leqq f( x)

And If \ \ \lim_{x\rightarrow a} f(x)\ \ =\ \ \lim_{x\rightarrow a} g(x) =\ P \\ \\
then, \ \ \ \lim_{x\rightarrow a} g(x)\ =\ P

## Questions on Sandwich Theorem of Limits

(01) Two functions are given
f(x) = 4x – 9
g(x) = x^{2} -4x+7

with the following condition:
f( x) \leq h( x) \leq g( x)

Find \lim _{x\rightarrow 4} h( x)

Solution
let\ us\ find\ \lim_{x\rightarrow 4} f( x)\\\ \\ \Longrightarrow \ \lim_{x\rightarrow 4} \ 4x-9\\\ \\ \Longrightarrow \ 4( 4) -9\\\ \\ \Longrightarrow \ 7 \\\ \\ Hence,\ \lim _{x\rightarrow 4} f( x) \ =7\\\ \\ \\\ \\ \\\ \\ Now\ find\ \ \lim_{x\rightarrow{}4}g(x) \\\ \\ ⟹ \lim_{x\rightarrow{}4} x^{2}-4x+7 \\\ \\ ⟹ (4)^{2}-4(4) +7 \\\ \\ ⟹ 7 \\\ \\ Hence, \lim_{x\rightarrow{}4}g(x) =7\\\ \\ \\\ \\ \\\ \\

From the above calculation, you can observe
\ \lim_{x\rightarrow 4} f( x) =\ \lim_{x\rightarrow 4} g( x) =7\

Then using the Sandwich Theorem,
\lim _{x\rightarrow 4} h( x) =\ 7

(02) Three functions follows following condition
f( x) \leq h( x) \leq g( x) \\\ \\
f(x) =\frac{-1}{4} x^{2} -\frac{1}{2} x\\\ \\ g( x) \ =\frac{1}{3} x^{2} +\frac{2}{3} x+\frac{2}{3}

Find \lim _{x\rightarrow -1} h(x)

Solution
Let us find the value of \lim _{x\rightarrow -1} f(x) \\\ \\ \Longrightarrow \lim _{x\rightarrow -1}\frac{-1}{4} x^{2} -\frac{1}{2} x\\\ \\ Putting\ x=-1\\\ \\ \Longrightarrow \frac{-1}{4}( -1)^{2} -\frac{1}{2}( -1)\\\ \\ \Longrightarrow \ -\frac{1}{4} +\frac{1}{2}\\\ \\ \Longrightarrow \ \frac{1}{4}\\\ \\ \\\ \\

Now find the value of \lim _{x\rightarrow -1} g(x)\\\ \\ \Longrightarrow \lim _{x\rightarrow -1}\frac{1}{3} x^{2} +\frac{2}{3} x+\frac{2}{3}\\\ \\ Putting\ x=-1\\\ \\ \Longrightarrow \ \frac{1}{3}( -1)^{2} -\frac{2}{3} +\frac{2}{3}\\\ \\ \Longrightarrow \ \frac{1}{3}\\\ \\

You can observe that f(x) is not equal to g(x) for limit x⟹-1, this means that both the function are not squeezing at a common point. So we cannot determine the value of function h(x) as the condition for sandwich theorem is not satisfied

(03) Find the value of following limit using Squeeze Theorem

Solution
Understand that the value of {x} always lies between 0 and 1

The above expression satisfy the condition for sandwich theorem

Where\ f( x) =\lim_{x\rightarrow \infty }\frac{0}{x}\ \\\ \\ g( x) \ =\lim_{x\rightarrow \infty }\frac{1}{x}\\\ \\ h( x) \ =\ \lim_{x\rightarrow \infty }\frac{{x}}{x}\\\ \\ and\\\ \\ f( x) \leq h( x) < \ g( x)\\\ \\ \\\ \\ \\\ \\ Solving\ f( x)\\\ \\ \Longrightarrow \ \lim_{x\rightarrow \infty }\frac{0}{x}\\\ \\ \Longrightarrow \frac{0}{\infty }\\\ \\ \ \Longrightarrow \ 0\ \ \\\ \\ \\\ \\ Solving\ g( x)\\\ \\ \Longrightarrow \lim_{x\rightarrow \infty }\frac{1}{x}\\\ \\ \Longrightarrow \ \frac{1}{\infty }\\\ \\ \Longrightarrow \ 0 \\\ \\

You can observe that f(x) = g(x) = 0 for limit x approaching infinity.
Now using the sandwich theorem, one can conclude that:

(04) It is given that
-3 ≤ f(x) < 5

find the following limit using Squeeze Theorem
\lim _{x\rightarrow 2} \ \left(( x-2)^{2} \ f( x) +1\right)\\\ \\ \\\ \\

Solution

-3\leq f( x) < 5 \\\ \\ Multiplying\ ( x-2)^{2}\\\ \\ -3( x-2)^{2} \leq ( x-2)^{2} f( x) < ( x-2)^{2} \ \times 5\\\ \\ \\\ \\ Adding\ +1\ \\\ \\ 1-3( x-2)^{2} \leq 1+( x-2)^{2} f( x) < 1+5( x-2)^{2} \\\ \\ \\\ \\ The\ equations\ follows\ sandwich\ rule\\\ \\ Solving\ 1st\ part\ of\ inequality\\\ \\ \Longrightarrow \ \lim_{x\rightarrow 2} \ \ \ 1-3( x-2)^{2}\\ \\ \Longrightarrow \ 1-3( 2-2)^{2}\ \ \Longrightarrow \ 1\\\ \\ \\\ \\ Solving\ 3rd\ part\ of\ inequality\\\ \\ \Longrightarrow \ \lim_{x\rightarrow 2} \ \ 1+5( x-2)^{2}\\\ \\ \Longrightarrow \ 1+5( 2-2)^{2}\\\ \\ \Longrightarrow \ 1\\\ \\ Since\ both\ the\ function\ is\ equal\ to\ 1\\\ \\ So,\ as\ per\ sandwich\ theorem\\\ \\ \lim _{x\rightarrow 2} \ \ 1+( x-2)^{2} f( x) \ =\ 1\\\ \\ Hence\ 1\ is\ the\ right\ answer\

(05) Find the limit of following Function
\lim _{x\rightarrow 0} \ \ x^{4} \times \ sin\left(\frac{3}{x}\right)

Solution
For conceptual clarity, let us try to solve this question with basic method

Put x =0 in the given function
⟹ 0^{4}× sin(\frac{3}{0}) \\\ \\ sin(\frac{3}{0})\ is\ undefined

Hence, we can’t solve the limit with this technique.
Let us now try to find the solution using Sandwich Method

We know that, the value of sin function lies between -1 and 1
-1\leq sin\left(\frac{3}{x}\right) \leq 1\\\ \\ Multiplying\ the\ equation\ with\ x^{4}\\\ \\ -x^{4} \leq sin\left(\frac{3}{x}\right) \leq x^{4}\\\ \\ \\\ \\ Putting\ the\ limit\ in\ the\ equation\\\ \\ \lim_{x\rightarrow 0} -x^{4} \leq \lim_{x\rightarrow 0} sin\left(\frac{3}{x}\right) \leq \lim_{x\rightarrow 0} x^{4}\\\ \\ \\\ \\ \ This\ is\ now\ taking\ the\ form\ of\ sandwich\ theorem\\\ \\ f( x) \leq h( x) \leq g( x)\\\ \\ \\\ \\ \ \ Here\ \ f( x) =-x^{4}\\\ \\ h( x) \ =sin\left(\frac{3}{x}\right)\\\ \\ g( x) \ =x^{4}\\\ \\ \\\ \\ \ \ Solving\ the\ limit\ of\ f( x)\\\ \\ \Longrightarrow \lim {x\rightarrow 0} -x^{4}\\\ \\ \Longrightarrow \ 0\\\ \\ \\\ \\ Solving\ the\ limit\ of\ g( x)\\\ \\ \Longrightarrow \lim {x\rightarrow 0} x^{4}\\\ \\ \Longrightarrow \ 0\\\ \\ \\\ \\ \ Since\ both\ f( x) \ =\ g( x)\\\ \\ Then\ using\ the\ sandwich\ theorem\\\ \\ \lim_ {x\rightarrow 0} h( x) \ =\lim _{x\rightarrow 0} sin\left(\frac{3}{x}\right) =0\