User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

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

Let $\lambda \in \R_{\ne 0}$ be 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:


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

Let $\lambda > 0$.

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

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

So:


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

So by 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 we can write $\lambda f = - \left({- \lambda f}\right)$ and invoke Negative of Function that Increases Without Bound.