# 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 $$x>0$$ 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 $$x<0$$ 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.

Let's start with the more famous problem, finding the limit of \frac{\sin x}{x}. Since $$\frac{\sin x}{x}$$ is even, we need only verify $$\lim_{x\to 0^+}\frac{\sin x}{x}=1$$. Draw a small angle $$x$$ between two length-$$1$$ radii $$OA,\,OB$$ of a centre-$$O$$ circle $$\gamma$$. Extend $$OA$$ through $$A$$ to meet the tangent-at-$$B$$ $$\ell$$ to $$\gamma$$ at $$C$$. The vertices $$O,\,A,\,B$$ are those of an area-$$\frac{1}{2}\tan x$$ triangle and an area-$$\frac{x}{2}$$ sector, and the right-angled $$\triangle OAC$$ has area $$\frac{1}{2}\tan x$$. Thus $$0\le\frac{\sin x}{x}\le 1\le\frac{\tan x}{x}.$$Multiplying by $$\cos^{\pm 1}x$$ (which is positive), $$0\le\cos x\le\frac{\sin x}{x}\le 1\le\frac{\tan x}{x}\le\sec x.$$(For example, get $$\cos x\le\frac{\sin x}{x}$$ from $$1\le\frac{\tan x}{x}$$.) Since $$\cos x\to 1$$, the squeeze theorem finishes the proof.

Now for the problem at hand. We've subtract at most $$|\sin x|$$ from the function considered above, but $$\sin x\to 0$$.

• The OP is about the integer part [1/x].
– user
Dec 6, 2018 at 18:48
• @gimusi My answer gets to that at the end because the first draft was written before that was clarified.
– J.G.
Dec 6, 2018 at 18:55

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

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

and then use squeeze theorem.