# Evaluating limit of indeterminate form

I am trying to formally evaluate the following limit: $\lim \limits_{x \rightarrow \infty} x \; \sqrt[]{\frac{3}{2}-{\frac{3x}{4\pi}}\sin(\frac{2\pi}{x})}$. Empirically, the function appears to be converging to $\pi$. Although I am wondering how to obtain this. Directly substituting $\infty$ for $x$ in the expression (dropping the $\frac{3}{2}$ in the radical) gives $\infty*\sqrt[]{-\infty*0}$, which I take is of indeterminate form. How might I use L'Hopital's Rule or another method to evaluate this limit?

Moving the $x$ into the radical gives us the following:
\begin{align} x\sqrt{\frac32-\frac{3x}{4\pi}\sin(\frac{2\pi}x)} & = \sqrt{x^2\left(\frac32-\frac{3x}{4\pi}\sin(\frac{2\pi}x)\right)} \\ & = \sqrt{\frac32x^2-\frac{3x^3}{4\pi}\sin(\frac{2\pi}x)} \\ \end{align}
consider the substitution $x=1/u$ to transform the limit into $\lim_{u\to0}$ and make the sine part in the more normal form. Then change all of the $\sin(u)$ to $u$, which is true for very small $u$.
$$E^2=\frac{3x^2}{2}-\frac{3x^3}{4\pi}\sin\left(\frac{2\pi}{x}\right)$$ Putting$\frac{2\pi}{x}=t$ we get $$E^2=\left(\frac{2\pi}{t}\right)^2\left(3-\frac 32\cdot\frac{\sin t}{t}\right)\Rightarrow \lim_{t\to 0}\left(\frac{2\pi}{t}\right)^2\left(3-\frac 32\cdot\frac{\sin t}{t}\right)=\infty$$ The expression $E$ tends to $\infty$.