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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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