Question #11a56

3 Answers
Feb 17, 2017

To find that the limit is #0#, I would reason as follows.

Explanation:

If #x# is very near #5#, then #absx# is very near #abs5# which is #5#.

So, if #x# is very near #5#, then #absx-5# is very near #abs5-5# which is #0#.

Feb 17, 2017

As #x# is positive around #x=5# then:

#lim_(x->5) (absx-5) = lim_(x->5) (x-5) = 0#

Explanation:

Consider the limit:

#lim_(x->5) (absx-5) = L#

what we need to prove is that for any #epsilon > 0# we can choose a #delta_epsilon > 0# such that:

#abs(x-5) < delta_epsilon => abs ( abs (x) - 5 -L) < epsilon#

Now if we choose for any #epsilon# a #delta_epsilon < 5#, we have that:

#abs(x-5) < delta_epsilon => 0 < 5-delta_epsilon < x < 5+delta_epsilon#

In other words, #x# is positive and #absx = x#

Thus:

#lim_(x->5) (absx-5) = lim_(x->5) (x-5) = 0#

In general, if we have:

#lim_(x->x_0) f(x)#

to determine the limit we only need to analyze the behavior of the function around #x_0#. So there are three cases:

1) #x_0 > 0#

In this case we can restrict the analysis to an interval where #x>0# and in that interval #abs x = x#

2) #x_0 < 0#

In this case we can restrict the analysis to an interval where #x<0# and in that interval #abs x = -x#

3) #x_0 = 0#

In this case in any interval containing #x_0# there are positive and negative values of #x#, but we know that if the limit exists then:

#lim_(x->0) f(x)#

must coincide with the right and left limit:

#lim_(x->0) f(x) = lim_(x->0^+) f(x) =lim_(x->0^-) f(x)#

But #x->0^+# means #x > 0# so #absx =x# and viceversa #x->0^-# means #x < 0# so #absx =-x#

Feb 17, 2017

Adding Socratic graph showing the limit #x to +-5_(+-) = 0#.

Explanation:

The other two answers have given details.

The graph here,for #y = |x|-5#, depicts that the limit is 0, either way #x to +-5#..
graph{|x|-5 [-12, 12, -6, 6]}