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

Dec 23, 2014

The answer is ${e}^{2}$.

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

Therefore, ${\left(1 + 2 x\right)}^{\frac{1}{\sin} x} = {e}^{u}$, where
$u = \ln \left({\left(1 + 2 x\right)}^{\frac{1}{\sin} x}\right) = \ln \frac{1 + 2 x}{\sin} x$

Therefore, as ${e}^{x}$ is continuous function, we may move limit:
${\lim}_{x \to 0} {e}^{u} = {e}^{{\lim}_{x \to 0} u}$

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 $\frac{0}{0}$.
${\lim}_{x \to 0} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 0} \left(\frac{f ' \left(x\right)}{g ' \left(x\right)}\right)$
Therefore,
${\lim}_{x \to 0} \ln \frac{1 + 2 x}{\sin} x = \frac{2}{2 x + 1} / \cos \left(x\right) = \frac{2}{\left(2 x + 1\right) \cos x} = 2$

And then, if we return to the original limit ${e}^{{\lim}_{x \to 0} u}$ and insert 2, we get the result of ${e}^{2}$,