# Homework Help: How to do this limit?

1. Mar 23, 2010

### zeion

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

$$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - sinx}$$

2. Relevant equations

3. The attempt at a solution

Do I need to do a l'hopital?

2. Mar 23, 2010

### Dick

Why are you asking? It's 0/0 form, isn't it?

3. Mar 24, 2010

### zeion

What do I do with a $$e^{-0}$$? Is that the same as just 0

4. Mar 24, 2010

### Dick

Sure. -0 is the same as 0. What's e^0?

5. Mar 24, 2010

### zeion

1
Ok so I do some l'hops and I get this

$$\frac {e^x + e^{-x} - x}{1 - cosx} = \frac {e^x - e^{-x} - 1}{sinx} = \frac {1-1-1}{1} = -1$$

6. Mar 24, 2010

### Dick

Look at your first step. Since when is the derivative of 2x equal to x?

7. Mar 24, 2010

### Staff: Mentor

Just to clarify, e-0 = e0 = 1. I believe Dick understood what you meant, zeion, even though you didn't ask the question clearly. IOW e-0 is not the same as 0.

8. Mar 24, 2010

### zeion

Ok sorry so I get like
$$\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?$$

9. Mar 24, 2010

### elect_eng

I feel like I'm reading a book called "Mathematical Calculations in the Twilight Zone".

10. Mar 24, 2010

### Dick

That's the right answer, but elect_eng makes a good point. Those quantities AREN'T equal. The LIMITS as x->0 of those quantities are equal. That's what makes it spooky to read.