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Limit involving exponential function question

  • Thread starter Pyroadept
  • Start date
  • #1
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Homework Statement


Find the limit as x tends to zero of: (e^-x - cos x)/2x


Homework Equations


lim_x->0 e^-x = 1
lim_x->0 cos x = 1
lim_x->0 sin x / x= 1


The Attempt at a Solution


Hi everyone,
Here's what I've done so far:

(e^-x - cosx)/2x = [(e^-x)^2 - (cosx)^2] / 2x(e^-x - cosx) .... multiplying by conjugate

= [e^-2x - 1 + (sinx)^2 ] / 2x(e^-x - cosx)
... And so I want to try and isolate the sinx to put it over the x, which will then go to 1. But I don't know how to do this. Or am I going about it the wrong way entirely?

Thanks for any help
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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I don't think that is going to work. You could taylor expand both the exponential and the cosine. This allows you to find the limit pretty easily.
 

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