posted by Anonymous .
lim as x-> 1
(x^2 + |x -1| - 1 ) / ( |x-1| )
Evaluate the limit approaching 1 from both x>1 and x<1 directins. The x-1 terms will get switched in sign in the latter case.
Use L'Hopital's rule and see if the limit is the same in both directions.
For x-> 1 from above, the limit is 2x + 1 = 3