How do we go from $\lim\limits_{x\to \infty} x\sin{\frac{1}{x}}$ to $\lim\limits_{x\to 0^+} \frac{1}{x}\sin{x}$? From Spivak's Calculus, Ch. 15 "Trigonometric Functions"


*Find $\lim\limits_{x\to \infty} x\sin{\frac{1}{x}}$.

The solution manuals says simply
$\lim\limits_{x\to \infty} x\sin{\frac{1}{x}}=\lim\limits_{x\to 0^+} \frac{1}{x}\sin{x}=1$
What is the exact manipulation at the $\epsilon$ $\delta$ level that allows us to change the form of the limit?
$\lim\limits_{x\to \infty} x\sin{\frac{1}{x}}=l$ means
$$\forall \epsilon>0\ \exists M>0\ \forall x, x>M\implies |x\sin{1/x}-l|<\epsilon$$
Let $y=\frac{1}{x}>0$.
Then for some $\epsilon>0$ there is an $M>0$ such that $\forall y$ such that $y=\frac{1}{x}$ we have
$$\left (y=\frac{1}{x}>M>0\implies 0<x<\frac{1}{M} \right )\implies \left |y\sin{\frac{1}{y}}-l \right |<\epsilon\implies \left |\frac{1}{x}\sin{x}-l \right |$$
Therefore, we are saying that for any $x$ we can form a $y$ larger than $M$, and this happens when $x$ is smaller than $1/M$ and this leads to $\left |\frac{1}{x}\sin{x}-l \right |$.
Therefore,
$$\forall \epsilon>0\ \exists \delta>0\ \forall x, 0<x<\delta \implies \left |\frac{1}{x}\sin{x}-l \right |$$
This sort of thing really makes my head spin. Is it really this difficult to grasp such manipulations? Are the manipulations of the initial limit above correct?
At this point we have
$$\lim\limits_{x\to 0^+} \frac{1}{x}\sin{x}=l$$
And this limit we can solve by L'Hôpital's Rule
$$\lim\limits_{x\to 0^+} \frac{1}{x}\sin{x}=\lim\limits_{x\to 0^+} \frac{\cos{x}}{1}=1$$
 A: I propose that instead of focusing on $x \sin x^{-1}$, we focus on a general function $f$.
We wish to show that if $\displaystyle\lim_{x\to\infty} f(x) = L$, then $\displaystyle\lim_{x\to0^+} f\left(\frac1x \right) = L$. The former gives us that for every $\varepsilon > 0$, there is some $N > 0$ such that $|f(x) - L| < \varepsilon$ whenever $x > N$. Similarly, we want to show that for every $\varepsilon > 0$, there is some $\delta > 0$ such that $|f\left( \frac1x \right) - L| < \varepsilon$ whenever $0 \le x < \delta$. This is equivalent to saying that if $0 \le \frac1x < \delta$, then $|f(x) - L| < \varepsilon$. Since $1/x < \delta$, we have that $x > \frac1\delta$. Therefore, choose $\delta = \frac1N$. Right now, we have shown that $\lim_{\frac1x \to 0^+} f(x) = L$, but let $y = \frac1x$, and we get $\lim_{y \to 0^+} f\left(\frac{1}{y}\right) = L$, and we are done.
A: The most didactic (in my opinion)  is the following
$$\lim_{t\to \infty}t\,\sin (t^{-1})=\lim_{t\to \infty}{\sin (t^{-1})\over t^{-1}}\ \underset{x=t^{-1}}{=}\ \lim_{x\to 0^+}{\sin x\over x}=1$$
More formally, the composition rule has been applied. Namely if $\displaystyle \lim_{x\to a^+}f(x)=b,$ $\displaystyle\lim_{t\to \infty}g(t)=a$ and $g(t)>a$ then
$$\lim_{t\to \infty}f(g(t))=b$$
In our case $g(t)=t^{-1},$ $f(x)=x^{-1}\sin x$ and $a=0,$ $b=1.$ I leave $\displaystyle\lim_{t\to \infty}t^{-1}=0$ as an easy fact.
The fastest way of proving the composition rule is by applying the Heine definition of the limit.
A: Consider functions $f(x)$ and $f(1/x)$ depicted below

The little arrow tips are just my way of showing where limits are being taken.
We have
$$\lim\limits_{x\to \infty} f(x)=L \tag{1}$$
$$\lim\limits_{x\to 0^+} f(1/x)=L\tag{2}$$
$(1)$ means
$$\forall \epsilon>0\ \exists N>0\ \forall x, x>N>0\implies |f(x)-L|<\epsilon\tag{3}$$
For a given $\epsilon>0$ we are looking at values of $x$ above a certain $N$.
If instead of writing a number as $x$ we write it as a multiplicative inverse $x^{-1}$ of some number $x$, then we have
$$0<N<\frac{1}{x}\implies 0<x<\frac{1}{N}\implies \left |f\left (\frac{1}{x}\right )-L\right |<\epsilon\implies |g(x)-L|<\epsilon \tag{4}$$
We can interpret $(4)$ in two ways

*

*for numbers $x^{-1}$ above $N$, $f$ evaluated at such numbers is close to $L$


*for numbers $x$ smaller than $\frac{1}{N}$, the function $g$ is close to $L$
