1. ## Hyperbolic Limit Question

I need to show that $\frac {\cosh(w)} {w^3}$ goes to zero as $w$ goes to infinity. I don't need a rigorous epsilon delta proof. Any help would be appreciated!

2. Originally Posted by Kep
I need to show that $\frac {\cosh(w)} {w^3}$ goes to zero as $w$ 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.

3. 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

4. Using L'Hospital's Rule:

$\lim_{w \to \infty}\frac{\cosh{w}}{w^3} = \lim_{w \to \infty}\frac{\sinh{w}}{3w^2}$

Since this is still of the indeterminate form $\frac{\infty}{\infty}$, use L'Hospital's rule again...

$\lim_{w \to \infty}\frac{\sinh{w}}{3w^2} = \lim_{w \to \infty}\frac{\cosh{w}}{6w}$.

This is still $\frac{\infty}{\infty}$, so use L'Hospital again.

$\lim_{w \to \infty}\frac{\cosh{w}}{6w} = \lim_{w \to \infty}\frac{\sinh{w}}{6}$.

This clearly tends to $\infty$ as $w \to \infty$.

5. Originally Posted by Kep
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
Yes, I noticed the same thing. There is a moral to the story ....

6. It is interesting that if you click the "show steps" button on the wolfram site, there are no steps, just a repeat of the claim that the limit is 0!

7. 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 $\mod( \frac {\cot(z)} {z^4})$ on the square with vertices $((N+\frac {1} {2})\pi)(\pm 1 \pm i)$

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...

8. Originally Posted by Kep
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 $\mod( \frac {\cot(z)} {z^4})$ on the square with vertices $((N+\frac {1} {2})\pi)(\pm 1 \pm i)$

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...
What is mod? Modulus?

9. Yeah, mod is modulus. As in $\sqrt(x^2+y^2)$ where $x=\Re(\frac {\cot(z)} {z^4})$ and $y=\Im(\frac {\cot(z)} {z^4})$