Talk:Push Theorem

Could this be proved as the contraposition of the Squeeze Theorem? --prime mover 12:37, 6 February 2012 (EST)
 * We don't have a page for the squeeze theorem for infinite limits of continuous functions...do we? --GFauxPas 12:47, 6 February 2012 (EST)

I'd say we can't, the Squeeze Theorem only takes care of finite intervals (at least for the moment). Maybe it can be expanded. --Lord_Farin 13:28, 6 February 2012 (EST)
 * I hope it can be expanded. When we were asked in Calc II to solve something like this, don't remember it exactly:


 * $\displaystyle \lim_{x \to \infty} \ \frac {\sin (x + 1)}{x}$ I used:


 * $-1 < \sin(x+1) < 1$


 * $\implies \dfrac {-1} {x} < \dfrac {\sin (x + 1)}x < \dfrac 1 x$


 * and took the limits of both sides, and it worked. My prof. could have been wrong, but my intuition is saying she's right. (Side note: Actually, I first used a trig identity, but I thought of this one later. She expected us to say that $\frac 1 x$ approaches $0$ and $\sin(1 + x)$ is bounded. ) --GFauxPas 13:45, 6 February 2012 (EST)