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Solution :

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

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