I was looking at limits involving $\sin x$ and was wondering if there exists a function $f(x)$ such that $$ \lim_{x \rightarrow \infty} \frac{f(x)}{\sin(\frac{a}{x})}=b, $$
where $$b \ne 0$$ $$\frac{df(x)}{da}=0$$ and $$\frac{db}{da}=0$$
In other words, is it possible for this to converge to a nonzero value that does not involve $a$ while also keeping $f(x)$ from involving $a$. Because one can do
$$ \lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{\sin(\frac{a}{x})}=\frac{1}{a}, $$ or $$\lim_{x \rightarrow \infty} \frac{\frac{a}{x}}{\sin(\frac{a}{x})}=1,$$
but $a$ has to appear either in $f$ or in the result. Is there a function where this doesn't happen? Also, it seems to me that
$$\lim_{x \rightarrow \infty} e^{-x}\frac{\frac{1}{x}}{\sin(\frac{a}{x})}=0,$$
and the only way to make this converge is to bring it to zero? Decided to ask because I could be missing some interesting limit where the above requirements are met.