Question

What is the limit of this ? `lim_(x->0) (e^x - e^-x -2x)/(x-sinx)`

Answer 1

`lim_(x->0) (e^x-e^-x-2x)/(x-"sen"(x)) = 2`

Explanation:

`lim_(x->0) (e^x-e^-x-2x)/(x-"sen"(x))`

La regla de L'Hospital:

`lim(f(x)/g(x)) = lim((f'(x))/(g'(x)))`

Usando la regla:

`lim_(x->0) (e^x+e^-x-2)/(1-"cos"(x))`

Otra vez:

`lim_(x->0) (e^x-e^-x)/("sen"(x))`

Otra vez:

`lim_(x->0) (e^x+e^-x)/("cos"(x))`

`= (e^(0) + e^(0))/("cos"(0))`

`=(1+1)/(1)`

`=2`

Answer 2

The limit equals `2`.

Explanation:

We immediately see that if we try and evaluate, we get `0/0`. Therefore, we may apply l'Hospitals.

`L = lim_(x-> 0) (e^x + e^-x - 2)/(1 - cosx)`

Once again trying to evaluate, we get `0/0`. L'Hospitals is going to have to be applied AGAIN!

`L = lim_(x-> 0) (e^x -e^(-x))/(sinx)`

Once again `0/0`. L'hospitals will be used once more.

`L =lim_(x-> 0) (e^x + e^(-x))/cosx`

Finally something we can evaluate through substitution!

`L = (e^0 + e^0)/cos(0) = 2/1 = 2`

Hopefully this helps!