Some Doubts in Limits from FIITJEE package

Ans 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.

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

No the ans given is ∞ ie (c)

Also how can u say that n^1-a is ∞ as 1-a lies between (0,1) as
a E (0,1)

@ Rajat
Here, a lies b/w o and 1 (given), therefore, 1-a > 0 and thus n ^1-a → ∞ as n → ∞