Why isn't $\lim_{x\to 0} x\sin(1/x)$ equal to $1$? How $\lim_{x\to 0} x\sin(1/x)$ is not equal to $1$?
$$
\lim_{x\to 0}  x\sin(1/x)  = \lim_{x\to 0}  \frac{\sin(1/x)}{1/x}=1
$$
What did I do wrong? My book says the answer is $0$.
But I used $\lim (\sin x )/ x=1 $ (formula).
 A: As $x$ approaches $0$, $\frac{1}{x}$ increases without bound. That is
$$
\lim_{x\to 0}\frac{1}{x} = \pm \infty
$$
(the $\pm$ depends on what side you approach $0$ from.)
Now $\sin$ of anything is always bounded between $-1$ and $1$. So, when you multiply this by something ($x$) that approaches $0$, the whole thing is going to approach $0$.
Your book probably tells you that
$$
\lim_{\color{red}{y\to 0}} \frac{\sin(y)}{y} = 1
$$
The limit you have is basically
$$
\lim_{t\to \infty} \frac{\sin(t)}{t} = 0.
$$
A: One way to think of it is $-1 \le \sin (1/x) \le 1$ so IF the $\lim_{x\rightarrow 0} x\sin(1/x)$ exists, then $\lim_{x\rightarrow 0} -x \le \lim_{x\rightarrow 0} x\sin(1/x) \le \lim_{x\rightarrow 0} x$. so $0 \le \lim_{x\rightarrow 0} x\sin(1/x) \le 0$.
So $\lim_{x\rightarrow 0} x\sin(1/x)= 0$.
A: I think you are confused with $$\lim_{x \rightarrow 0} \frac{sin(x)}{x}=1$$
$$\lim_{x \rightarrow \infty} \left|\frac{sin(x)}{x}\right|\leq \lim_{x \rightarrow \infty} \left|\frac{1}{x}\right|=0$$
A: check this page for formatting help
MathJax basic tutorial and quick reference
$\lim_\limits {x\to 0} x \sin \frac 1x = 0$
Why?
$-1\le \sin \frac 1x\le1\\
-x\le x\sin \frac 1x\le x$
And as $x$ goes to $0, x\sin \frac 1x$ goes to $0$ by the squeeze theorem.
What have you done?
$\lim_\limits {x\to 0} x \sin \frac 1x$
Suppose we replace $x$ with $\frac 1x$
$\lim_\limits {x\to 0} x \sin \frac 1x = \lim_\limits{x\to \infty} \dfrac {\sin x}{x}$
All well and good, but then you evaluated the limit on the right as it approached 0, and not as it approached infinity.
