Question

`lim_(x->0)(e^x-1)/sin(2x)=` ?

Answer 1

It diverges.

Explanation:

`lim_(x->0) e^x - 1/sin (2x)` can be written as
`lim_(x->0) e^x - lim_(x->0) 1/sin(2x)`.
First limit is equal to `1`.
The second limit approaches infinity, because
`sin(2x) = 2sinx*cosx`, and
`lim_(x->0) (2sinx*cosx) = 0`.

So `lim_(x->0) sin^(-1) (2x)` approaches infinity.

Then the original limit is just `1-"infinity" = "- infinity"`.

Answer 2

`1/2`

Explanation:

Assuming that the question reads

`lim_(x->0)(e^x-1)/sin(2x)` we have

`e^x = 1+x+O(x^2)`
`sin(2x) = 2x-(2x)^3/(3!)+O(x^5)` and then

`lim_(x->0)(e^x-1)/sin(2x)= lim_(x->0)(x+O(x^2))/( 2x-(2x)^3/(3!)+O(x^5)) = 1/2`