Negative of Real Function that Increases Without Bound
Jump to navigation
Jump to search
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 infinite limits 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 negative infinite limit at infinity.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analagous.
$\blacksquare$