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## Homework Statement

[tex]\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - sinx}[/tex]

## Homework Equations

## The Attempt at a Solution

Do I need to do a l'hopital?

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- Thread starter zeion
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- #1

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[tex]\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - sinx}[/tex]

Do I need to do a l'hopital?

- #2

Dick

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

- #3

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What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0

- #4

Dick

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What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0

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

- #5

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

[tex]\frac {e^x + e^{-x} - x}{1 - cosx} = \frac {e^x - e^{-x} - 1}{sinx} = \frac {1-1-1}{1} = -1[/tex]

- #6

Dick

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

[tex]\frac {e^x + e^{-x} - x}{1 - cosx} = \frac {e^x - e^{-x} - 1}{sinx} = \frac {1-1-1}{1} = -1[/tex]

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

- #7

Mark44

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What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0

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

Just to clarify, e

- #8

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[tex]\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?[/tex]

- #9

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[tex]\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?[/tex]

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

- #10

Dick

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[tex]\frac {e^x + e^{-x} -2}{1-cosx} = \frac {e^x - e^{-x}}{sinx} = \frac {e^x + e^{-x}}{cosx} = 2?[/tex]

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.

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