limits doubt..

i hope samajh main a rha hai..

sorry , i cannot understand the question

plzzz see 18th question......i hope nw its clear

x1 = 2cos(pi/12)

x2 = 2cos(pi/24)

xn = 2cos(pi/(6*2ⁿ))

now

√2-xn = 2sin(pi/(6*2ⁿ⁺¹))

so it simplifies to

lt 2ⁿ⁺²sin(pi/(6*2ⁿ⁺¹))
n->∞

kinowing as A->0 sinA->A ..................

2ⁿ⁺²pi/(6*2ⁿ⁺¹)

= pi/3 !!!!!!!!!!!!!

forgive calc errors!!!!!!!!!!!!!!