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.

  • 2
    $\begingroup$ Since $f(-x) = -f(x)$ and negation is continuous, then yes. $\endgroup$
    – copper.hat
    Oct 29, 2015 at 5:27
  • 3
    $\begingroup$ If either limit exists then so does the other and they are the same. However, the limits may not exist. The continuity part has no bearing on the question. $\endgroup$ Oct 29, 2015 at 5:27
  • $\begingroup$ Suppose the limits are infinite. $\endgroup$ Oct 29, 2015 at 5:29
  • 2
    $\begingroup$ Things are OK also if we allow $\pm \infty$ as limits. But for example if $f(x)=\sin x$ then neither limit exists. $\endgroup$ Oct 29, 2015 at 5:33
  • $\begingroup$ Great. That agrees with my previous intuition. Thanks for the prompt responses. $\endgroup$ Oct 29, 2015 at 5:37

2 Answers 2


Let's say:


It means that:

$$\forall \epsilon>0,~~~~~ \exists M>0, x>M \Rightarrow |f(x)-L|<\epsilon$$

Since $f(-x)=-f(x)$

$$\forall \epsilon>0,~~~~~ \exists M>0, x>M \Rightarrow |-f(-x)-L|<\epsilon$$


$$\forall \epsilon>0,~~~~~ \exists -M<0, x<-M \Rightarrow |f(-x)-(-L)|<\epsilon$$

$$\forall \epsilon>0,~~~~~ \exists N<0, x<N \Rightarrow |f(-x)-(-L)|<\epsilon$$




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 $x \to \infty$.

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