Question #df591

The #epsilon-delta# definition of a limit states that given a function #f:D->RR# (a function with the domain #D# and codomain #RR#), then we say that #f# has a limit of #L# at a value #a# if for for every #epsilon > 0# there exists a #delta > 0# such that for any #x in D# where #0 < |x - a| < delta#, we have #|f(x) - L| < epsilon#.

In other words, #lim_(x->a)f(x)=L# if we can make #f(x)# arbitrarily close to #L# by choosing values of #x# from the domain of #f# which are close to #a#.

Using this, we can show that #lim_(x->0)-sqrt(x) = 0#.

Proof: Let #epsilon > 0# be arbitrary. Let #delta = epsilon#. Then, for any #x in [0, oo)#, if #|x - 0| < delta#, we have

#|-sqrt(x) - 0| = |-sqrt(x)| = |sqrt(x)| <= x = |x - 0| < delta = epsilon#.

Thus, by the above definition, with #f(x) = -sqrt(x)#, and #a = L = 0#, we have #lim_(x->0)-sqrt(x) = 0#.

For the definition of #lim_(xrarra)f(x)=L#, if we begin by requiring

#f# is defined (and real valued) on some open interval containing #a# except possibly at #x = a#,

then #lim_(xrarr0)(-sqrtx)# is not defined.

For a limit from the right, #lim_(xrarra^+) f(x) = L#, we simply required that #f# is defined on some interval of the form #(a,b)#.

In this case #lim_(xrarr0^+) (-sqrtx) # is defined and is #0#.

If one uses the definition cited by sente in that answer, then, as shown, #lim_(xrarr0) (-sqrtx) = 0#