$\lim x \sin (1/x)$, when $x \to 0$ Here's my solution: $$\lim x\sin (1/x) = \lim\, x \dfrac{\sin (1/x)}{x(1/x)} = \lim\, x/x = 1$$ when $x \to 0$
However on the internet I read that the solution of this equation is 0. How can this be? Where a I making a mistake?
 A: Notice that:
$0\le|x\sin(\frac{1}{x})|\le|x|$
so the limit is $0$. In your solution you wrote:
$\lim_{x\to0}\frac{x\sin(\frac{1}{x})}{\frac{x}{x}}=\lim_{x\to0}\frac{x}{x}$.
I'm afraid I don't see why this is true.
A: I think you are trying to assume that $$\lim_{x \to 0}\dfrac{\sin(1/x)}{(1/x)} = 1$$ where as in reality we have $$\lim_{x \to 0}\frac{\sin x }{x} = 1$$ and this $x$ in $(\sin x) / x$ can be replaced by any other expression (like $x^{2}, \sin x$ etc) which tends to zero when $x \to 0$. Thus we may write $$\lim_{x \to 0}\frac{\sin(\sin x)}{\sin x} = 1$$
In your case you have committed an error in replacing $x$ by $1/x$. Note that when $1/x$ does not tend to $0$ as $x \to 0$.
A: Hint:
$$
\lim_{x\to0}x\sin\left(\frac{1}{x}\right) = \lim_{t\to\infty}\frac{\sin(t)}{t}
$$
Now, try to think why your solution is not true.
A: Well see function $\sin\left(\frac{1}{x}\right)$ is bounded between $[-1,1]$  and as $$\lim_{x \to 0}x\sin\left(\frac{1}{x}\right)$$ the value of  $x$ will also go to zero hence value of limit is $0$.
A: This solution is not true because the formula is : $\lim_{x\to 0} \frac{\sin(x)}{x} =1$ whereas here there is $\sin(1/x)$ so this can not be applied , $\lim \frac{\sin(1/x)}{1/x}$ will be $1$ when $\frac{1}{x} \to 0$.
