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

Question

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 .

Answer 1

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.

Answer 2

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 $\sin x\to 0$.

Answer 3

HINT

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

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

then assuming wlog $0<x<1$

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

and then use squeeze theorem.