Elementary Infinite Limit Question Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous odd function. Does the following statement hold (assuming the limits exist): 
$$ \lim_{x\to-\infty}f(x)=-\lim_{x\to\infty}f(x)$$
This certainly seems to be true to me, because we can make this statement for any arbitrarily large, finite $x$. However, I just wanted to ensure that this statement holds in the infinite case.
 A: Let's say:
$$\lim_{x\to~+\infty}f(x)=L$$
It means that:
$$\forall \epsilon>0,~~~~~ \exists M>0, x>M \Rightarrow |f(x)-L|<\epsilon$$
Since $f(-x)=-f(x)$
$$\forall \epsilon>0,~~~~~ \exists M>0, x>M \Rightarrow |-f(-x)-L|<\epsilon$$
or
$$\forall \epsilon>0,~~~~~ \exists -M<0, x<-M \Rightarrow |f(-x)-(-L)|<\epsilon$$
$$\forall \epsilon>0,~~~~~ \exists N<0, x<N \Rightarrow |f(-x)-(-L)|<\epsilon$$
or
$$\lim_{x\to~-\infty}f(x)=-L$$
A: Clearly $$\lim_{x \to -\infty}f(x) = \lim_{y \to \infty}f(-y) = \lim_{y \to \infty}(-f(y)) = -\lim_{x \to \infty}f(x)$$ The above holds without any regard to continuity of $f$ and also without any regard to the existence of the limit.
However, when the limit does not exist we need to interpret the equality in a different manner. Thus we have the following options:


*

*If $\lim_{x \to -\infty}f(x)$ exists then $\lim_{x \to \infty}f(x)$ also exists and $\lim_{x \to -\infty}f(x) = -\lim_{x \to \infty}f(x)$.

*If $f(x) \to \infty$ as $x \to -\infty$ then $f(x) \to -\infty$ as $x \to \infty$.

*If $f(x) \to -\infty$ as $x \to -\infty$ then $f(x) \to \infty$ as $x \to \infty$.

*If $f(x)$ oscillates finitely as $x \to -\infty$ then $f(x)$ also oscillates finitely as $x \to \infty$.

*If $f(x)$ oscillates infinitely as $x \to -\infty$ then $f(x)$ also oscillates infinitely as $x \to \infty$.

