Question #581eb

We will most likely be using L'hopital's rule here. L'hopital's rule states that provided that both the numerator and denominator are at least locally differentiable and continuous, then in situations where #lim_(x->c) # is indeterminate (e.g. of the form #0/0 or (+-oo)/(+-oo)#...

#lim_(x->c) (f(x))/(g(x)) = lim_(x->c) ((df)/(dx))/((dg)/(dx)) =lim_(x->c) (f')/(g')#

We will call #ae^x+e^(-x) + bx= f(x), and 1-cosx= g(x)#
The denominator, #1-cos(x)# , will go to 0 as we approach 0 for x #( 1 - cos(0) = 1-1 = 0)#, whereas for the top ...

#f(0) = ae^(0) +e^(-0)+b(0) = a + 1#

In order for this to reach indeterminate form, then, we set this equal to 0...

#a+1 = 0 -> a = -1#

And find our value for a.

Now, we differentiate the top and bottom, recalling that we know #a=-1#...

#lim_(x->0) f/g = lim_(x->0) (f')/(g') =lim_(x->0) (-e^x - e^(-x) + b)/(sin x)#

Again, the denominator goes to 0 as x approaches 0, so we must make the numerator go to 0 as well.

#-e^(0) - e^(-0) + b = 0 -> -1 -1 + b = 0 -> b-2 = 0 -> b = 2#

Thus, for the limit to exist, our original expression must be:

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