Negative of Real Function that Increases Without Bound

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

Then:


 * $(1): \quad \displaystyle \lim_{x \mathop \to +\infty} f \left({x}\right) = +\infty \implies \lim_{x \mathop \to +\infty} -f \left({x}\right) = -\infty$


 * $(2): \quad \displaystyle \lim_{x \mathop \to -\infty} f \left({x}\right) = +\infty \implies \lim_{x \mathop \to -\infty} -f \left({x}\right) = -\infty$

Proof
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 $f \left({x}\right) > M \iff -f \left({x}\right) < -M$.

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.

Also see

 * Negative of Real Function that Decreases Without Bound