Tuesday, 5 November 2019

calculus - Evaluation of limnrightarrowinftyleftleft(2+sqrt3right)2nright;, Where ninmathbbN.

Evaluation of lim Where n\in \mathbb{N}.




\bf{My\; Try::} Let \left(2+\sqrt{3}\right)^{2n} = I +f\;, where $0

Now Let 0<\left(2-\sqrt{3}\right)^{2n}<1\;, So \left(2-\sqrt{3}\right)^{2n}=f'.



So \left(2+\sqrt{3}\right)^{2n}+\left(2-\sqrt{3}\right)^{2n} = I +f+f' = \bf{Integer\; Quantity.}



So f+f'\in \mathbb{Z}. Now Given 0<(f+f')<2. So f+f' = 1\in \mathbb{Z}



So I+f+f' = \bf{Integer\; Quantity}\Rightarrow f = Integer\; Quantity-f'




So \displaystyle \lim_{n\rightarrow \infty}\left\{\left(2+\sqrt{3}\right)^{2n}\right\} = \bf{Integer\; quantity-}\displaystyle \lim_{n\rightarrow \infty}\left\{\left(2-\sqrt{3}\right)^{2n}\right\}



Now how can i solve after that, help me, Thanks

No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...