I need to understand why the limit of $x\cdot \sin (1/x)$ as $x$ tends to infinity is 1 here's the question, how can I solve this:
$$\lim_{x \rightarrow \infty} x\sin (1/x) $$
Now, from textbooks I know it is possible to use the following substitution $x=1/t$, then, the ecuation is reformed in the following way 
$$\frac{\sin t}{t}$$
then, and this is what I really can´t understand, textbook suggest find the limit as $t\to0^+$ (what gives you 1 as result)
Ok, I can't figure out WHY finding that limit as $t$ approaches $0$ from the right gives me the answer of the limit in infinity of the original formula. I think I can't understand what implies the substitution.
Better than an answer,  I need an explanation.
(Sorry If I wrote something incorrectly, the english is not my original language)
              Really thanks!!
 A: $$\lim_{x\to\infty}x\sin(1/x)$$ is like the limit of this sequence: $$1\sin(1/1), 10\sin(1/10), 100\sin(1/100),\ldots$$ where we have inserted $x$ marching along from $1$ to $10$ to $100$ on its merry way to $\infty$. This is almost literally the same as $$\frac{1}{1}\sin(1), \frac{1}{0.1}\sin(0.1), \frac{1}{0.01}\sin(0.01),\ldots$$ which is an interpretation of $$\lim_{t\to0^+}\frac{1}{t}\sin(t)$$ with $t$ marching its way from $1$ down to $0.1$ down to $0.01$ on its merry way to $0$. So whatever the value of these limits are, $\lim\limits_{x\to\infty}x\sin(1/x)=\lim\limits_{t\to0^+}\frac{1}{t}\sin(t)$.
A: You can simply observe that
$$
\lim_{x\rightarrow+\infty}x\sin(1/x)=
\lim_{x\rightarrow+\infty}x\frac{1}{x}\frac{\sin(1/x)}{\frac{1}{x}}=
\lim_{x\rightarrow+\infty}\frac{\sin(1/x)}{1/x}
$$
Now it should be clear that $1/x$ tends to $0$ as $x$ approaches to $+\infty$. Hence it will be equivalent to write $1/x=t$ and let $t$ tends to $0$.
Thus you have
$$
\lim_{x\rightarrow+\infty}\frac{\sin(1/x)}{1/x}=
\lim_{t\rightarrow0}\frac{\sin t}{t}=1\;,
$$
as wanted.
