Doubt from calculus

Question no 64

What's the answer

I am getting a what is the answer

Think by some \textit{trignometric angle} i think its value of some trignometric function may be tan or cot 18^° not confirmed

Mthd 2 go with binomial where we have to find integral part by manipulating imagine
(2+\sqrt{3})^n=I+f
Dont have pen paper so not able to solve but think question demand is this mthd

Let we take (2+\sqrt{3})^n=I+f\cdots\cdots (1)

and let take (2-\sqrt{3})^n=f'\cdots\cdots\cdots(2)

Where 0 \leq f<1 and 0<f'<1.

Then (2+\sqrt{3})^n+(2-\sqrt{3})^n=\text{even} integer.

So we have I+f+f'=\text{even} integer.

\Longrightarrow f+f'\in\mathbb{Z}. from above 0<f+f'<2.

\Longrightarrow f+f'=1\Longrightarrow f=1-f'.

So \lim_{n\rightarrow \infty}\{(2+\sqrt{3})^n\}=1-\lim_{n\rightarrow \infty}(2-\sqrt{3})^n=1.

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Answer is 1 correct

Yes same method @abhijit_2020