Negative of Real Function that Decreases Without Bound

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

Then:


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


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

Proof
Suppose $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty$.

Then by the definition of negative infinite limit at infinity:


 * $\forall M < 0: \exists N > 0: x > N \implies \map f x < M$

But:
 * $M < 0 \iff -M > 0$

Likewise:
 * $\map f x < M \iff -\map f x > -M$

Putting $M' = -M$:


 * $\forall M' > 0: \exists N > 0: x > N \implies -\map f x > M'$

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

The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous.

Also see

 * Negative of Real Function that Increases Without Bound