This in an exercise in my Analysis book in a section on L'Hopital's rules. $$\lim\limits_{x \to \infty} \left[\sin\left(\frac{1}{x}\right)\right]^x$$ Now it's an indeterminate of the form $0^\infty$ however I don't know how to solve this. I have tried the following:
$$y=\lim\limits_{x \to \infty} \left[\sin\left(\frac{1}{x}\right)\right]^x$$ $$\ln y=\lim\limits_{x \to \infty}\ln \left[\sin\left(\frac{1}{x}\right)\right]^x$$ $$\ln{y}=\lim\limits_{x \to \infty} x\ln{\sin\frac{1}{x}}$$ Now this is an indeterminate limit of form $\infty\cdot\infty$ which approaches $\infty$. However I may not write now that therefore $y=e^\infty=\infty$.
How do I write this out correctly?