Negative of Real Function that Increases Without Bound

Theorem
Let $f: \R \to \R$ be a real function.

Then:


 * $\displaystyle \lim_{x \mathop \to +\infty} f\left({x}\right) = +\infty \iff  \lim_{x \mathop \to +\infty} -f\left({x}\right) = -\infty$


 * $\displaystyle \lim_{x \mathop \to -\infty} f\left({x}\right) = +\infty \iff  \lim_{x \mathop \to +\infty} -f\left({x}\right) = -\infty$

Sufficient Condition
Suppose $\displaystyle \lim_{x \mathop \to +\infty} f\left({x}\right) = +\infty$.

Then by the definition of infinite limits at infinity:


 * $\forall M > 0: \exists N > 0: x > N \implies f\left({x}\right) > M$.

But $M > 0 \iff -M < 0$.

Likewise $f\left({x}\right) > 0 \iff -f\left({x}\right) < 0$.

Putting $M' = -M$:


 * $\forall M > 0: \exists N > 0: x > N \implies f\left({x}\right) > M$.

The result then follows from the definition of infinite limits at infinity.

The proof for $\displaystyle \lim_{x \mathop \to -\infty} f\left({x}\right) = +\infty$ is analagous.

Necessary Condition
As $-\left({-f\left({x}\right)}\right) = f\left({x}\right)$, the result follows follows from the sufficient condition.

Also see

 * Negative of Function that Decreases Without Bound