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
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.


Answer is 1 correct

Yes same method @abhijit_2020