I was doing an exercise in limits and I was working with the following limit:
$$\lim_{y\to\infty}{y\sin{\frac{1}{y}}}$$
I understand that I’m meant to use the identity
$$\lim_{x\to 0}{\frac{\sin{x}}{x}}=1$$
to reduce this somehow, and so I reasoned that I could take $y=x^{-1}$ and get
$$\lim_{x^{-1}\to\infty}{\frac{\sin{x}}{x}}$$
I’m almost there, but I’m not sure how to convert the conditions of the limit. I know that $x^{-1}\to\infty \iff x \to 0$, but am I allowed to simply make the jump and say
$$\lim_{x^{-1}\to\infty}{\frac{\sin{x}}{x}} =\lim_{x\to0}{\frac{\sin{x}}{x}}$$
just like that? Is there a theorem or axiom or definition that vindicates this step?
Let me bring one theorem from Kudriavtsev L.D. "Course of Mathematical Analysis", 4.8*, 1 tom. 1981, page 108 (I couldn't find the answer I posted before, probably deleted).
Assume exists finite or infinite limits $\lim\limits_{x\to a}f(x)=b$ and $\lim\limits_{y\to b}F(y)$. Also let's assume, that exists some deleted neighborhood of $a$ in which holds $f(x)\ne b$. Then exists limit of compound function $F\circ f$ and holds
$$\lim\limits_{x\to a}F(f(x))=\lim\limits_{y\to b}F(y)$$
In your case $F(y) = y\sin\frac{1}{y}$ and $y=f(x)=\frac{1}{x}$. I hope it will not be difficult for you to check the conditions of application.