11Ans 1) lim n→∞ n a sin 2 (n!)n+1
= n a sin 2 (n!)n(1 + 1/n)
= sin 2 (n!)n 1-a (1+1/n)
= a finite quantity∞ = 0
[As sin 2 (n!) always lie b/w 0 and 1. Also, since 1-a > 0 , therefore, n 1-a → ∞ as n → ∞ ]
Therefore, (a) is the correct answer.
11
Tush Watts
·2009-12-28 22:42:08
Ans 2) Put 1/ sin 2 x = t
Therefore, lim n→∞ (1 t + 2 t + 3 t + ..............+ n t) 1/t
= lim n→∞ (n t) 1/t [ (1/n) t + (2/n) t + ...............+ 1] 1/t
= n lim n→∞ [ (1/n) t + (2/n) t + ...............+ 1] 1/t
= n [0 + 0 + ........+1] 0 = n
Ans 3) lim x→∞ x n + nx n-1 +1e [x]
Now put x = a+k , where n E I , 0 ≤ k ≤ 1 ; as x → ∞ , a → ∞
Then, lim x→∞ x n + nx n-1 +1e [x] = lim a→∞ (a+k) n + n (a+k) n-1 + 1e [a+k]
= lim a→∞ (a+k) n + n (a+k) n-1 + 1e a
1No the ans given is ∞ ie (c)
Also how can u say that n1-a is ∞ as 1-a lies between (0,1) as
a E (0,1)
11
Tush Watts
·2009-12-28 22:50:07
@ Rajat
Here, a lies b/w o and 1 (given), therefore, 1-a > 0 and thus n 1-a → ∞ as n → ∞