Limit of (e^x - e^-x - 2x)/(x - sinx) | How To Solve

[zeion]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Do I need to do a l'hopital?

 

[Dick]

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

 

[Dick]

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

 

[Dick]

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?

 

[Mark44]

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?

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".

 

[Dick]

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?

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.