Negative of Real Function that Decreases Without Bound
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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.
$\blacksquare$