How do you find the limit of #| x - 5 |# as x approaches #5#?

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Answer 1 · 2016-10-13T12:05:22

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#abs(x-5)# is a continuous funtion so #lim_(x to 5) abs ( x-5) = abs (5 - 5) = 0#

Answer 2 · 2016-10-13T18:56:52

If you are doing this to prove that the function is continuous, rewrite using the definition of absolute value.

#absu = {(u," if ",u >= 0),(-u," if ",u < 0) :}#

With #u = x-5#, the condition #u >= 0# becomes #x-5 >= 0# which is equivalent to #x >= 5#.
Similarly #u < 0#, becomes #x < 5#.

So,

#abs(x-5) = {(x-5," if ",x >= 5),(-(x-5)," if ",x < 5) :}#

#lim_(xrarr5^+)abs(x-5) = lim_(xrarr5^+)(x-5) = 0#
and
#lim_(xrarr5^-)abs(x-5) = lim_(xrarr5^-)(-(x-5)) = -(0)=0#

Since both one-sided limits are #0#, the limit is #0#.