# Value of Cosh - 1 / h

1. Jun 2, 2013

### physphys

1. The problem statement, all variables and given/known data

Use a calculator to evaluate the quantity (cosh-1) / h for h = 0.1, 0.01, 0.001, and 0.0001.
Estimate the value of the limit of h -> 0 of (cos h -1) / h

2. Relevant equations

3. The attempt at a solution
okay so at 0.1 it is 1.523 x 10^-5 and it goes on like that
I'm in Degree mode but I don't think it'll matter.
I know the derivative of cosx is -Sinx but for some reason when I plug the numbers in, they don't match. Please help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 2, 2013

### SteamKing

Staff Emeritus
Is your function [cos(h-1)]/h? If you don't use parentheses properly, cos h can be confused with the hyperbolic cosine (cosh).

And yes, if it is [cos (h-1)]/h, you probably should be using radians instead of degrees for your calculations, particularly when evaluating limits.

3. Jun 2, 2013

### physphys

I think it's cos(h-1) the question has it written the way I wrote it. H is the same as ΔX because it is a limit. Okay, but is my assumption that the value should be equivalent to sin correct?

4. Jun 2, 2013

### Ray Vickson

Do you mean
$$\frac{\cosh(-1)}{h},\\ \cosh\left(\frac{-1}{h}\right),\\ \frac{\cos(h-1)}{h}, \text{ or}\\ \cos \left( \frac{h-1}{h}\right)?$$
What you wrote is most consistent with the *first* interpretation, but the others can also be inferred reasonably. Obviously, you need to use parentheses, like this: cosh(-1)/h or cosh(-1/h) or cos(h-1)/h or cos[(h-1)/h].

5. Jun 2, 2013

### Mute

Based on what the OP is tasked with doing, I'm guessing the correct interpretation is

$$\frac{cos(h)-1}{h},$$
which appears in the calculation of the derivative of $\sin(x)$.

6. Jun 2, 2013

### physphys

yes that's what I meant, $$\frac{cos(h)-1}{h},$$
I know that it is in fact the derivative of sinx but when I input the values it tells me in the calculator I'm not getting the same.

7. Jun 2, 2013

### Dick

No, it's not the derivative of sin(x), it's a difference quotient for the derivative of cos(x) at x=0. What does your calculator tell you?

8. Jun 2, 2013

### LCKurtz

Of course, this is just part of the calculation for the derivative of sin(x). Your original post looks like it shows that this fraction goes to zero. That's what it should do. What about the rest of the calculation?