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```        find
7 years ago

105 Points
```										Dear Baha,
AS limit x-> 0+
mod(x+1) is always positive
and mod(x-1) = (1-x) because for a small nighobur hood in 0+ the value of x-1 comes out to be negative
Hence for limit x->0+
hte functions is =  (x+1-(1-x))/x = 2
As limit x->0-
mod(x+1) is always positive for a samll neighbour hood in 0- x+1 is alwyas positive
mod(x-1) is = -(x-1)
Hence for limit x->0-
((x+1)- (-(x-1))) /x = 2
All the best.
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```
7 years ago
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