Scaled Real Function that Decreases Without Bound

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

Let $\lambda \in \R_{\ne 0}$ be a nonzero constant.

Then:

For $\lambda > 0$:


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


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

For $\lambda < 0$:


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


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

Proof
Let $\displaystyle \lim_{x \mathop \to +\infty} f\left({x}\right) = -\infty$.

From the definition of infinite limits at infinity, this means that:


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

Suppose $\lambda > 0$.

Then $M < 0 \iff \lambda^{-1}M < 0$.

Also, $f \left({x}\right) < \lambda^{-1}M \iff \lambda f\left({x}\right) < M$.

So:


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

From the definition of infinite limits at infinity:


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

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

Now, suppose $\lambda < 0$.

Then $-\lambda f < 0$, and so $-\lambda f \to -\infty$, from above.

Write $\lambda f = - \left({- \lambda f}\right)$ and the result follows from Negative of Real Function that Decreases Without Bound.

Also see

 * Scaled Real Function that Increases Without Bound