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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)

```
8 years ago

147 Points
```										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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```
8 years ago
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