I need to show that $\displaystyle \frac {\cosh(w)} {w^3}$ goes to zero as $\displaystyle w$ goes to infinity. I don't need a rigorous epsilon delta proof. Any help would be appreciated!
The limit is not zero. The limit is actually +oo.
One proof of this limit is to divide the Maclaurin series for cosh(w) by w^3 and then consider the limit. Another proof is to use l'Hopital's Rule since it's an indeterminant form oo/oo. A simple minded approach is to substitute larger and larger values of w into the expression and observe the pattern.
That's what I had thought... but Wolfram Alpha seems to think that it is zero...
Limit of cosh(w)/w^3 as w goes to infinity - Wolfram|Alpha
Using L'Hospital's Rule:
$\displaystyle \lim_{w \to \infty}\frac{\cosh{w}}{w^3} = \lim_{w \to \infty}\frac{\sinh{w}}{3w^2}$
Since this is still of the indeterminate form $\displaystyle \frac{\infty}{\infty}$, use L'Hospital's rule again...
$\displaystyle \lim_{w \to \infty}\frac{\sinh{w}}{3w^2} = \lim_{w \to \infty}\frac{\cosh{w}}{6w}$.
This is still $\displaystyle \frac{\infty}{\infty}$, so use L'Hospital again.
$\displaystyle \lim_{w \to \infty}\frac{\cosh{w}}{6w} = \lim_{w \to \infty}\frac{\sinh{w}}{6}$.
This clearly tends to $\displaystyle \infty$ as $\displaystyle w \to \infty$.
Wolfram Alpha: The limit is zero
Me: Really? Why? *Clicks show steps*
Wolfram Alpha: The limit IS zero dammit, stop being so curious...
This is bad news for me though
I'm supposed to find an upper bound for $\displaystyle \mod( \frac {\cot(z)} {z^4})$ on the square with vertices $\displaystyle ((N+\frac {1} {2})\pi)(\pm 1 \pm i)$
This is then supposed to vanish when multiplied by the perimeter of the square as N goes to infinity.
My upper bound times perimeter was of that form and since Wolfram Alpha was encouraging I assumed I had got it right...
Serves me right for not posting the whole question...
Any ideas? Sorry this might be a bit tough...
My first attempt was to write cot in terms of exp and use the triangle inequality...