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 $\lambda > 0$:
- $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
- $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = -\infty$
For $\lambda < 0$:
- $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$
- $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = +\infty$
Proof
Let $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty$.
From the definition of infinite limits at infinity, this means that:
- $\forall M < 0: \exists N > 0: x > N \implies \map f x < M$
Suppose $\lambda > 0$.
Then $M < 0 \iff \lambda^{-1}M < 0$.
Also, $\map f x < \lambda^{-1}M \iff \lambda \map f x < M$
So:
- $\forall M < 0: \exists N > 0: x > N \implies \lambda \map f x < M$
From the definition of infinite limits at infinity:
- $\ds \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous.
Now, suppose $\lambda < 0$.
Then $-\lambda f < 0$, and so $-\lambda f \to -\infty$, from above.
Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Decreases Without Bound.
$\blacksquare$