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.