Question

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

Answer 1

2

Explanation:

`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`