How do you find the limit of #(4-v)/abs(4-v)# as #v->4^+#?

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Answer 1 · 2016-10-19T16:33:14

#abs(4-v) = {(4-v," if ",4-v >= 0),(-(4-v)," if ",4-v < 0):}#

# = {(4-v," if ",4 >= v),(-(4-v)," if ",4 < v):}#

So,

#(4-v)/abs(4-v) = {((4-v)/(4-v)," if ",4 >= v),((4-v)/(-(4-v))," if ",4 < v):}#

#= {(1," if ",4 >= v),(-1," if ",4 < v):}#

As #vrarr4^+# we have #4 < v#, so

#lim_(vrarr4^+)(4-v)/abs(4-v) = lim_(vrarr4)(-1) = -1#