# How to do this limit?

zeion

## Homework Statement

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

## The Attempt at a Solution

Do I need to do a l'hopital?

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

$$\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?$$