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f(x) = |x| + |x-1| at x = 1....

`f(x) = |x| + |x-1|` at `x = 1`.

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The correct Answer is:
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We have, `f(x) = |x| + |x-1|` at `x = 1`.
At `x = 1, LHL = underset(xrarr1^(-))(lim)[|x|+|x-1|]`
`=underset(hrarr0)(lim)[|1+h|+|1-h-1|]=1+0=1`
and `RHL = underset(xrarr1^(-))lim[|x|+|x-1|]`
`=underset(hrarr0)(lim)[|1+h|+|1+h-1|]=1+0=1`
and `f(1)=|1|+|0|=1`
`:. LHL = RHL= f(1)`
Hence, `f(x)` is continuous at `x = 1`.
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