How does one prove this statement or provide a counterexample?

Is it possible to prove the following statement or provide a counterexample function for the following statement?
Thanks in advance.

#lim_(x->-oo)(f(x))=lim_(x->oo)(f(-x))#

1 Answer
Sep 7, 2017

It's true.

Explanation:

Technically, there are cases to be considered when #f rarr oo# and when #f rarr -oo#. I will illustrate a proof for the case when the limit on the left exists.
Assume #lim_(x rarr -oo) f(x) = L#.
Let #epsilon > 0#.
Since the limit is L, then there exists an M such that whenever #x < M#, we have #|f(x) - L| < epsilon#.

Now examine #f(-x)#. We will introduce a dummy variable, y.
Whenever #y > -M#, it is true that #-y < M#.
For such y, given the above information,
#|f(-y) - L| < epsilon#.

Let #N = -M#.
Then there exists an N such that when #y > N#, we have
#|f(-y) - L| < epsilon#.
This is the definition of the limit on the right side.