Question

What is the limit as x approaches 0 of `(1+2x)^cscx`?

Answer 1

The answer is `e^2`.

The reasoning is not that simple. Firstly, you must use trick: a = e^ln(a).

Therefore, `(1+2x)^(1/sinx) = e^u`, where
`u=ln((1+2x)^(1/sinx)) = ln(1+2x)/sinx`

Therefore, as `e^x` is continuous function, we may move limit:
`lim_(x->0) e^u = e^ (lim_(x->0)u)`

Let us calculate limit of `u` as x approaches 0. Without any theorem, calculations would be hard. Therefore, we use de l'Hospital theorem as the limit is of type `0/0`.
`lim_(x->0) f(x)/g(x) = lim_(x->0)((f'(x))/(g'(x)))`
Therefore,
`lim_(x->0) ln(1+2x)/sinx = 2/(2x+1)/cos(x) = 2/((2x+1)cosx) = 2`

And then, if we return to the original limit `e^ (lim_(x->0)u)` and insert 2, we get the result of `e^2`,