Find $$\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$$ where $\lfloor\cdot\rfloor$ is greatest integer function

I am a high school teacher. One of my students came up to ask this limit. For $\lfloor\frac{\sin x}{x}\rfloor$, I have used $\sin x > x$ using increasing decreasing functions.

I tried to prove $x^2 > \sin x \tan x$ using increasing /decreasing function but I am not getting it.

  • $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Jul 13 '13 at 16:39
  • $\begingroup$ am new to stack exchange it is limit x->o integral part of (x^2 /sinx.tanx) $\endgroup$ – raj Jul 13 '13 at 16:45
  • 1
    $\begingroup$ $\sin x < x$ if $x>0$ $\endgroup$ – Empy2 Jul 13 '13 at 16:49

Apply GM - HM to $\sin x$ and $\tan x$ (both positive for $x\geq 0$), we get that

$$ \sqrt{ \sin x \tan x } \geq \frac{2} { \frac{1} {\sin x} + \frac{ 1}{ \tan x} } = \frac{2 \sin x} { 1 + \cos x } = 2 \tan \frac{x}{2} \geq x$$

The only 'calc' that you need is the last inequality, though it has an easy geometric solution.

For $x < 0$, both $\sin x, \tan x$ have the same sign, and you can do the above with absolute values instead, no difference.

  • $\begingroup$ note that : $x\neq 0$ Because $\sin 0=\tan 0=0$ So $\dfrac{1}{\sin x},\dfrac{1}{\tan x} =\text{Undefined}$ $\endgroup$ – Almot1960 Jun 1 '17 at 7:27

Consider the function $$f(x) = \frac{x^2\cos(x)}{\sin^2(x)}$$ Then we have that $$f(x+h)\approx \cos(x) < 1$$ for every $h\neq 0$ with $|h|$ small enough. Thus the limit converges to $0$.

We want to show that $f(x)<1$ for $|x|\neq0$ small enough. We will do so by showing that $g(x) = \sin^2(x)>x^2\cos(x)=h(x)$. The two functions agree in $0$, thus it is enough to show that $g'(x)>h'(x)$ for $x>0$ (we don't have to consider $x<0$, since both $g$ and $h$ are even functions). $$g'(x) = 2\sin(x)\cos(x)$$ $$h'(x) = x\cos(x)-x^2\sin(x)$$ Since both agree in $0$ (again) we can consider the next derivative: $$g''(x) = 2(\cos^2(x) - \sin^2(x))$$ $$h''(x) = \cos(x) - x(3\sin(x) -x\cos(x))$$ Notice that $g''(0)=2>1=h''(0)$, thus by continuity of both functions there is a neighborhood of $0$ where $g''>h''$, and we're done.

  • $\begingroup$ Put my answer together with the one of Omnomnomnom to get a complete solution. $\endgroup$ – Daniel Robert-Nicoud Jul 13 '13 at 16:57
  • $\begingroup$ But $x^2/\sin^2x$ is more than 1, and it's not clear (certainly in high school maths) which dominates. $\endgroup$ – Empy2 Jul 13 '13 at 16:57
  • $\begingroup$ Is there a way to show that $\frac{x}{\sin x}\to 1$ faster than $\cos(x)$? $\endgroup$ – Ben Grossmann Jul 13 '13 at 16:58
  • $\begingroup$ @Omnomnomnom Looking at the Taylor expansion does the job: $\frac{\sin(x)}{x} = 1 - \frac{x^2}{6} + O(x^4)$, thus $\frac{\sin^2(x)}{x^2} = 1 - \frac{x^2}{3}+O(x^4)$, while $\cos(x) = 1-\frac{x^2}{2}+O(x^4)$. $\endgroup$ – Daniel Robert-Nicoud Jul 13 '13 at 17:04
  • $\begingroup$ I am high school level so my students know basics of limits, continuity and differentiability. is there any way without Taylor series, by simply proving that numerator is > or < denominator,such that floor function gives zero or one $\endgroup$ – raj Jul 13 '13 at 17:34

So first of all, note that $$ \lim_{x\to 0}\frac{x^2}{\sin x \tan x}=\dots= \left(\lim_{x\to 0}\frac{x}{\sin x}\right)^2 \left(\lim_{x\to 0}\cos(x)\right)=1 $$ Now, this doesn't necessarily tell us anything about the original function, since the greatest integer function is not continuous. So, we actually have three possibilities:

  • if $x^2\geq\sin x \tan x$ in a sufficiently small neighborhood of $x=0$, then the limit is $1$.
  • if $x^2<\sin x \tan x$ in a sufficiently small neighborhood of $x=0$, then the limit is $0$.
  • if neither of the above is true, the limit does not exist.

So, we must find out: which is it?

Here's an attempt to show that the limit becomes $0$: $$ \begin{align} x^2 &< \sin x \tan x\\ x^2\cos x &< \sin^2 x \\ x^2\cos x &<\frac12 (1-\cos(2x))\\ 2x^2\cos x &< (1-\cos(2x))\\ 2x^2\cos x + \cos(2x) &< 1\\ \end{align} $$ From there, it might be possible to make an argument using Taylor series.

  • $\begingroup$ thanks a lot , but i cannot use Taylor series $\endgroup$ – raj Jul 13 '13 at 17:36

Since you already know how get the limit without the floor function, I will try to prove that inequality without Taylor series.

$$x^2<\sin x \tan x \quad as \; x \to 0$$

I made the substitution $x \to \arctan x$ .

$\arctan^2 x<x\sin (\arctan x)$

$\arctan x < \large \frac{x}{(x^2+1)^{\frac 14}}$

There are two functions $f(x)$ and $g(x)$ . $f(0)=g(0)$ . If $f'(x)>g'(x)$ on the interval $(0, a)$ , then that implies that $f(x)>g(x)$ on the interval $(0, a)$ . Therefore if $RHS'>LHS'$ , then $RHS>LHS$ .

$\large \large \large \frac {1}{x^2+1} <\frac {x^2+2}{2(x^2+1)^{\frac 54}}$

$1<\large \frac {x^2+2}{2(x^2+1)^{\frac 14}}$

Using standard techniques (such as first derivative test) we can show that the $RHS$ has a minimum at $(0, 1)$ so we have proved the inequality. Hope this helps!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.