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1. Lt n→ ∞ sin 2 [π√((n!)2-n! )]
2. Ltx→0 ({x})1/x + (1/x){x} , (x>0,{.} denotes fractional part of x)
please solve both the Questions.
Dear sindhu
first question
Lt n→ ∞ sin 2 [π√((n!)2-n! )]
or Lt n→ ∞ sin 2 [∏n!√( 1-1/n! )]
apply limit part inside the root will become 1
and n! is very large nimber but it is a even number.
so Lt n→ ∞ sin 2 [∏n!√( 1-1/n! )] = sin 2 [∏*even number]
=0
For second question
Ltx→0 ({x})1/x + (1/x){x}
as limit tends to zero {x}=x
so Ltx→0 ({x})1/x + (1/x){x} =Ltx→0 (x)1/x + (1/x)x
apply limit
Ltx→0 (x)1/x + (1/x)x = 0 +1=1
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