Prove $\underset{x\to\infty}{\lim}\frac{\sin{(1/x})}{1/x}=1$ without L'Hospital's rule.

Question

I have the following problem:

Prove $\displaystyle\lim_{x\to\infty}\frac{\sin{(1/x})}{1/x}=1$ without L'Hospital's rule.

I tried to use the sandwiches theorem but it didn't work.
My attempt: $$-1\leq\sin\left(\frac{1}{x}\right)\leq1$$ $$-x\leq\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\leq x$$ $$\lim_{x\to\infty}{-x}=-\infty$$ $$\lim_{x\to\infty}{x}=\infty$$ $$-\infty\leq\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\leq \infty$$ That's where I got stuck.
Thank you very much for your precious time!

Answer 1

For $x\not=0$, consider the change of variables $y:=1/x$. If $x\to +\infty$, then $y\to 0^{+}$. So $$\lim_{x\to +\infty}\frac{\sin(1/x)}{1/x}=\lim_{y\to 0^{+}}\frac{\sin y}{y}=1.$$ Of course, you can prove the last limit without the L'Hôpital's rule, see here.