How to do this limit?

  • Thread starter zeion
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  • #1
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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?
 

Answers and Replies

  • #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
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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-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
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Ok sorry so I get like
[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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Ok sorry so I get like
[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". :smile:
 
  • #10
Dick
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Ok sorry so I get like
[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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