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How do you find the limit of #(e^x - e^-x - 2x) / (x - Sinx)# as x approaches 0?

Answer 1

#lim_(x->0) (e^x-e^-x-2x)/(x-sinx) = (e^0-e^-0-2(0))/(0-sin0) = (1-1-0)/(0-0) = 0/0#

This is an indeterminate type so use l'Hopital's Rule which is the limit of the quotient of the derivative of the top and the derivative of the bottom as x goes to 0.

#lim_(x->0) (e^x +e^-x-2)/(1-cosx) = (e^0+e^-0-2)/(1-cos 0) =(1+1-2)/(1-1) = 0/0#

Still an indeterminate form so apply l'Hopital's rule again

#lim_(x->0) (e^x-e^-x)/sinx = (e^0 -e^-0)/sin0 =( 1-1)/0 = 0/0#

Still an indeterminate form so apply l'Hopital's rule again

#lim_(x->0) (e^x+e^-x)/cos x = (e^0+e^-0)/cos 0 = (1+1)/1=2#

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Answer 2

Scarlett Allison

To find the limit of (e^x - e^-x - 2x) / (x - Sinx) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (e^x + e^-x - 2) / (1 - Cosx). Evaluating this expression at x = 0, we have (1 + 1 - 2) / (1 - 1) = 0. Therefore, the limit of the given expression as x approaches 0 is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.