Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $

Question

Find the limit

$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$

My try:

$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$

$$\sin 2x =2\cos x\sin x$$

So: $$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$

So:

$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor= \lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$

Now what?

An easy way if you know Taylor expansions: $\frac{\tan(2x)}{\sin x} = 2+3x^2 + o(x^2)$ when $x\to 0$, so $2< \frac{\tan(2x)}{\sin x} < 3$ for $x$ small enough. (Only posting that as a comment, can elaborate in an answer if you want.) — Clement C., May 30 '17 at 16:32
You can show that $\frac{\cos x}{\cos(2x)}$ takes the minimum $1$ at $0$. So in a neighborhood of $0$, your function is constant 2. — nrkm, May 30 '17 at 16:42

score 5 · Accepted Answer · answered May 30 '17 at 16:38

5

$x\neq0$ and around $0$ we have $$\frac{2\cos{x}}{\cos2x}>2$$ because it's $$\frac{2\cos{x}}{2\cos^2x-1}>2$$ or $$(1-\cos{x})(1+2\cos{x})>0.$$ Thus, $$\lim_{x\rightarrow0}\left[\frac{2\cos{x}}{\cos2x}\right]=2$$

answered May 30 '17 at 16:38

Michael Rozenberg

194,933

I think it is important to mention that the fraction is also less than $3$. Just showing that it is greater than $2$ is not sufficient. – Paramanand Singh May 31 '17 at 10:32
@Paramanand Singh it's obvious because $\lim\limits_{x\rightarrow0}\frac{2\cos{x}}{\cos2x}=2$. – Michael Rozenberg May 31 '17 at 11:58

score 4 · Answer 2 · answered May 30 '17 at 16:49

Let's consider $$ f(x)=\frac{2\cos x}{\cos2x} $$ over $(-\pi/2,\pi/2)$. Then $$ f'(x)=2\frac{-\sin x\cos2x+2\cos x\sin2x}{\cos^22x}= 2\frac{\sin x(2\cos^2x+1)}{\cos^22x} $$ Therefore $f$ has a local minimum at $0$ and $f(0)=2$.

In a suitable neighborhood of $0$ we have $2\le f(x)<3$.

score 2 · Answer 3 · answered May 30 '17 at 16:57

Note that

\begin{align} \frac{2\cos x}{\cos 2x}-2&=\frac{2\cos x-2\cos 2x}{\cos 2x}\\ &=\frac{4\sin\frac{3x}{2}\sin\frac{x}{2}}{\cos 2x}\\ &>0 \end{align}

for $x\in(\frac{-\pi}{4},\frac{\pi}{4})$.

We also have

\begin{align} 3-\frac{2\cos x}{\cos 2x}&=\frac{3\cos 2x-2\cos x}{\cos 2x}\\ &=\frac{6\cos^2x-2\cos x-3}{\cos 2x}\\ &=\frac{36\cos^2x-12\cos x-18}{6\cos 2x}\\ &=\frac{(6\cos^2x-1)^2-19}{6\cos 2x}\\ &>0 \end{align}

for $x\in(-\alpha,\alpha)$, where $\alpha=\arccos(\frac{1+\sqrt{19}}{12})$

Note that $0<\alpha<\frac{\pi}{4}$.

For $x\in(-\alpha,\alpha)$, $\displaystyle\lfloor \frac{2\cos x}{\cos 2x}\rfloor=2$.

The limit is $2$.

Paramanand Singh · Answer 4 · 2017-05-31T11:04:54.360

The function in question is even and hence it is sufficient to deal with $x\to 0^{+}$. And then we have the famous inequality $$\sin x<x<\tan x\tag{1}$$ so that $$\frac{\tan 2x}{\sin x} =\frac{\tan 2x}{2x}\cdot 2\cdot\frac{x}{\sin x} > 1\cdot 2\cdot 1=2$$ And clearly from the above equation it is evident that $(\tan 2x)/\sin x\to 2$ so that as $x\to 0^{+}$ we have $(\tan 2x)/\sin x<5/2$. It follows that the function $[(\tan 2x)/\sin x] $ is constant with value $2$ as $x\to 0^{+}$ and hence the desired limit is $2$.

Your approach is also fine and you may do without inequality $(1)$. You just have to note that as $x\to 0^{+}$ we have $x<2x$ and $\cos$ is decreasing so that $\cos x>\cos 2x$ and then $$\frac{2\cos x} {\cos 2x}>2$$ and the limit of the fraction is $2$ so that it is less than $5/2$ as $x\to 0^{+}$. And as before the desired limit is $2$. The question is much simpler than it appears.

Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $

4 Answers4