Prove that function hasn't lim in #x_0=0# ?

Function:
#f(x)={(2 " when " x=(1,1/2,1/4...)), (1 " when " x=R\(1,1/2,1/4...)):}#

2 Answers
Dec 18, 2017

See explanation.

Explanation:

According to Heine's definition of a function limit we have:

#lim_{x->x_0} f(x)=g iff #
#AA{x_n} (lim_{n->+oo}x_n=x_0 => lim_{n->+oo}f(x_n)=g)#

So to show that a function has NO limit at #x_0# we have to find two sequences #{x_n}# and #{bar(x)_n}# such, that

#lim_{n->+oo}x_n=lim_{n->+oo}bar(x)_n=x_0#

and

#lim_{n->+oo}f(x_n)!=lim_{n->+oo}f(bar(x)_n)#

In the given example such sequences can be:

#x_n=1/(2^n)# and #bar(x)_n=1/(3^n)#

Both sequences converge to #x_0=0#, but according to the function's formula we have:

#lim _{n->+oo}f(x_n)=2# (*)
because all elements in #x_n# are in #1,1/2,1/4,...#

and for #bar(x)_n# we have:

#f(bar(x)_1)=f(1)=2#

but for all #n>=2# we have: #f(bar(x)_n)=1#

So for #n->+oo# we have:

#lim_{n->+oo}f(bar(x)_n)=1# (**)

Both sequences coverge to #x_0=0#, but the limits (*) and (**) are NOT equal, so the limit #lim_{x->0}f(x)# does not exist.

QED

The limit definition can be found in Wikipedia at: https://en.wikipedia.org/wiki/Limit_of_a_function

Dec 18, 2017

Here is a proof using the negation of the definition of the existence of a limit.

Explanation:

Short version

#f(x)# cannot approach a single number #L# because in any neighborhood of #0#, the function #f# takes on values that differ from each other by #1#.

So no matter what someone proposes for #L#, there are points #x# near #0#, where #f(x)# is at least #1/2# unit away from #L#

Long version

#lim_(xrarr0)f(x)# exists if and only if

there is a number, #L# such the for all #epsilon > 0#, there is a #delta > 0# such that for all #x#, #0 < abs(x) < delta# implies #abs(f(x)-L) < epsilon#

The negation of this is:

#lim_(xrarr0)f(x)# fails to exist if and only if

for every number, #L# there is an #epsilon > 0#, such that for all #delta > 0# there is an #x#, such that #0 < abs(x) < delta# and #abs(f(x)-L) >= epsilon#

Given a number #L#, I will let #epsilon = 1/2# (any smaller #epsilon# will work as well)

Now given a positive #delta#, I must show that there is an #x# with #0 < absx < delta# and #abs(f(x)-L) >= 1/2# (recall that #epsilon = 1/2#)

Given a positive #delta# , eventually #1/2^n < delta# so there is an #x_1# with #f(x_1) = 2#.
There is also an element #x_2 in RR-{1, 1/2, 1/4, . . . }# with #0 < x_2 < delta# and #f(x_2) = 1#

If #L <= (1/2)#, then #abs(f(x_1)-L) >= 1/2#

If #L >= (1/2)#, then #abs(f(x_2)-L) >= 1/2#