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# ISI 2021 Objective Problem 23 I A Problem from Limit Try this beautiful Objective Limit Problem appeared in ISI Entrance - 2021.

## Problem

Let us denote the fractional part of

a real number $x$ by $\{x\}.$

(Note ${x}=x-[x]$ where $[x]$

is the integer part of $x$.)

Then
$\lim _{n \rightarrow \infty}{(3+2 \sqrt{2})^{n}}.$

(A) equals 0
(B) equals 1
(C) equals $\frac{1}{2}$
(D) does not exist

### Key Concepts

Fractional part of a real number

Greatest Integer function

## Suggested Book | Source | Answer

IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2021 , Objective problem number - 23

(B) equals 1

## Try with Hints

Hint 1

Try to find the fractional part of $(3+2 \sqrt{2})^{n} = N$(Let) .

Hint 2

$\bullet$ Observe $(3+2 \sqrt{2})^{n} + (3-2 \sqrt{2})^{n}$ is an integer.

$\bullet$ And use $0 < (3-2 \sqrt{2}) < 1.$

Hint 3

$\bullet$ Obviuosly $0<(3-2 \sqrt{2})^{n}<1.$

$\bullet$ Also assume , $p=(3-2 \sqrt{2})^{n}.$

Hint 4

$\bullet$ As $N+p = integer = [N] + \{N\} + p ,$

so $\{N\} + p = integer - [N]= integer.$

$\bullet$ Hence proceed.

Conclusion

$\bullet$ It is very easy to find that $\{N\} + p =1.$

∙ Therefore ,

$\lim _{n \rightarrow \infty}(3+2 \sqrt{2})^{n}$

=limn→∞{N}=limn→∞(1p)=1

[As,limn→∞p=limn→∞(3−22)n=0]

ISI Entrance Program at Cheenta

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Try this beautiful Objective Limit Problem appeared in ISI Entrance - 2021.

## Problem

Let us denote the fractional part of

a real number $x$ by $\{x\}.$

(Note ${x}=x-[x]$ where $[x]$

is the integer part of $x$.)

Then
$\lim _{n \rightarrow \infty}{(3+2 \sqrt{2})^{n}}.$

(A) equals 0
(B) equals 1
(C) equals $\frac{1}{2}$
(D) does not exist

### Key Concepts

Fractional part of a real number

Greatest Integer function

## Suggested Book | Source | Answer

IIT mathematics by Asit Das Gupta

ISI UG Entrance - 2021 , Objective problem number - 23

(B) equals 1

## Try with Hints

Hint 1

Try to find the fractional part of $(3+2 \sqrt{2})^{n} = N$(Let) .

Hint 2

$\bullet$ Observe $(3+2 \sqrt{2})^{n} + (3-2 \sqrt{2})^{n}$ is an integer.

$\bullet$ And use $0 < (3-2 \sqrt{2}) < 1.$

Hint 3

$\bullet$ Obviuosly $0<(3-2 \sqrt{2})^{n}<1.$

$\bullet$ Also assume , $p=(3-2 \sqrt{2})^{n}.$

Hint 4

$\bullet$ As $N+p = integer = [N] + \{N\} + p ,$

so $\{N\} + p = integer - [N]= integer.$

$\bullet$ Hence proceed.

Conclusion

$\bullet$ It is very easy to find that $\{N\} + p =1.$

∙ Therefore ,

$\lim _{n \rightarrow \infty}(3+2 \sqrt{2})^{n}$

=limn→∞{N}=limn→∞(1p)=1

[As,limn→∞p=limn→∞(3−22)n=0]

ISI Entrance Program at Cheenta

## Subscribe to Cheenta at Youtube

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