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You won't hardly get around standard limits.
Here is a way using the standard limits $\lim_{u\to 0}\frac{\sin u}{u} = 1$ and $\lim_{u\to 0}\frac{ln(1+t)}{t}=1$:
Using $\sin(\pi +u)=-\sin u$ and $x=1+t$ and considering $t\to 0^+$ you get
\begin{eqnarray*}\frac{\sin^2\pi x}{x\ln^2x}
& \stackrel{x=1...