1.

Discuss the continuity of `f(x)=|x|+|x-1| ` at x=0 and x=1.

Answer» `f(x)=|x|={:{(x, xge 0),(-x,x lt 0):}`
and `abs(x-1)={:{(x-1, xge 1),(-(x-1), xlt 1):}`
`therefore f(x)=abs(x)+|x-1|={:{(1-2x, x lt 0),(1, 0 le x lt1),(2x-1, x ge1):}`
At x=0
f(0)=1
`R.H.L=underset(xrarr0^+)lim=underset(xrarr)lim f(0+h)=underset(xrarr0)lim 1=1`
`L.H.L=underset(xrarr0^-)lim=underset(xrarr0)lim f(0+h)`
`=underset(hrarr0)lim{-(0-h)}=0`
`therefore` R.H.L = f(0)=L.H.L
`therefore` f(x) is constinuous at x=0 Hence proved.


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