Show that $g(x)=[\frac1x]\sin x$ has a limit in $x=0$. Show that $g(x)=\left[\frac1x \right]\sin x$ has a limit in $x=0$.

( $[1/x]$ as greatest integer less than or equal to $1/x$)

I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .

I know :
$$\frac{1}{x}-1\le\left[\frac{1}{x}\right]\le\frac{1}{x},x\not=0.$$

And I know :

$$\lim_{x\rightarrow 0} \frac{1}{x}\sin(x)=1$$

I just don’t know how to use this stuffs to solve the question.

Thanks in advance fo the help .
If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x&gt;0$ we have $$\left(\frac{1}{x}-1\right)\sin(x)≤[\frac{1}{x}]\sin(x)≤\frac{1}{x} \sin(x)$$ since $$\frac{1}{x}-1≤[\frac{1}{x}]≤\frac{1}{x},x\not=0.$$ Also for sufficiently small $x&lt;0$ we have $$\left(\frac1x-1\right)\sin(x)≥[\frac{1}{x}]\sin(x)≥\frac{1}{x} \sin(x).$$ Now $$\lim_{x\rightarrow 0} \frac{1}{x}\sin(x)=1$$ and $$\lim_{x\rightarrow0}\sin(x)=0.$$ Hence $$\lim_{x\rightarrow 0}[\frac1x]\sin(x)=1.$$

The function $\sin$ is positive for small positive numbers and negative for small negative numbers.
HINT

You are mostly done, indeed since by floor function definition we have

$$\frac1x-1\le\left[\frac1x\right]\le \frac1x$$

then assuming wlog $0&lt;x&lt;1$

$$ \left(\frac1x-1\right)\cdot\sin x\le\left[\frac1x\right]\cdot\sin x\le \frac1x\cdot\sin x$$

and then use squeeze theorem.