Limit involving exponential function question

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SUMMARY

The limit as x approaches zero for the expression (e^-x - cos x)/2x can be evaluated using Taylor series expansions for both the exponential function and the cosine function. The relevant limits are lim_x->0 e^-x = 1 and lim_x->0 cos x = 1, which simplifies the expression significantly. By applying Taylor expansions, one can isolate terms that lead to the limit being computed directly, confirming that the approach of using conjugates is less effective in this case.

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  • Understanding of limits in calculus
  • Familiarity with Taylor series expansions
  • Knowledge of exponential functions and trigonometric functions
  • Basic algebraic manipulation skills
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  • Study Taylor series expansions for e^x and cos x
  • Learn techniques for evaluating limits using L'Hôpital's Rule
  • Explore the properties of exponential and trigonometric functions
  • Practice solving limits involving indeterminate forms
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Students studying calculus, particularly those focusing on limits and series expansions, as well as educators looking for effective teaching strategies in limit evaluation.

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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
 
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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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