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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?
Sure. -0 is the same as 0. What's e^0?What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0
Look at your first step. Since when is the derivative of 2x equal to x?1
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]
What do I do with a [tex]e^{-0}[/tex]? Is that the same as just 0
Just to clarify, e^{-0} = e^{0} = 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.Dick said:Sure. -0 is the same as 0. What's e^0?
I feel like I'm reading a book called "Mathematical Calculations in the Twilight Zone".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.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]