How do you evaluate # (5-t) / | 5- t | # as x approaches #5^-#?

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Answer 1 · 2016-05-25T11:50:15

Hence #t->5^-# that means #5-t>0# hence #abs(5-t)=5-t#

and #lim_(t->5^-) (5-t)/(abs(5-t))=lim_(t->5^-) (5-t)/(5-t)=1#

Answer 2 · 2016-05-25T13:32:28

The function is discontinuous at t = 5. The limit #= +-1#

#|5-t|=5-t, when t < 5 and t-5, when t > 5.

So, for #t < 5, (5-t)/|5-t|= (5-t)/(5-t)=1

and for any #t >5, (5-t)/|5-t|= (5-t)/(t-5)=-1#.

The answer is now clear, #+-1#.

and the graph for # x = (5-t)/|5-t|# reveals the fall, at t = 5.