Question

Question #581eb

Answer 1

See explanation. `a=-1, b=2`

Explanation:

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)`