$\int_0^\infty ne^{-nx}\sin\left(\frac1{x}\right)\;dx\to ?$ as $n\to\infty$

I want to find limit of $\displaystyle\int_0^\infty ne^{-nx}\sin\left(\frac{1}{x}\right)\;dx$ as $n\to\infty$ if it exists or to prove that it doesn't exist. I see that $ne^{-nx}\sin\left(\frac{1}{x}\right)\to 0$ for all $x>0$ and that the convergence is uniform on $[a,\infty)$ for all $a>0$. That implies $\displaystyle\int_a^\infty ne^{-nx}\sin\left(\frac{1}{x}\right)\;dx\to 0$ as $n\to\infty$ for all $a>0$. Can anyone tell me what the next step is or if I'm on the wrong track? Thanks.

Change variables $y = nx$. Your integral becomes $$\int_0^{\infty}e^{-y}\sin({n \over y})\,dy$$ $$= {{1 \over n}}\int_0^{\infty}y^2e^{-y}{{n \over y^2}}\sin({n \over y})\,dy$$ Integrate this by parts, integrating ${n \over y^2}\sin({n \over y})$ to $\cos({n \over y})$, and differentiating $y^2e^{-y}$. One obtains $${1 \over n}\lim_{y \rightarrow \infty} \big(y^2 e^{-y}\cos({n \over y})\big) - {1 \over n}\lim_{y \rightarrow 0} \big(y^2 e^{-y}\cos({n \over y})\big) - {1 \over n}\int_0^{\infty}(2y - y^2)e^{-y}\cos({n \over y})\,dy$$ Observing that both the boundary terms go to zero, this becomes $$- {1 \over n}\int_0^{\infty}(2y - y^2)e^{-y}\cos({n \over y})\,dy$$ Since $|\cos({n \over y})| \leq 1$ for all $n$ and all $y$, the above is bounded in absolute value by $${1 \over n}\int_0^{\infty}(2y + y^2)e^{-y}\,dy$$ Since the integral doesn't depend on $n$ and is finite, the limit as $n$ goes to infinity of the above expression is zero.