Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ [duplicate]

Question

Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.

My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function.

So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$

Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion

So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$

Now how can i solve after that, Help me

Thanks

Answer 1

Hint?

$$ \lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x} = \lim_{x\to 0} \cos x \left( \frac{x}{\sin x} \right)^2 $$

Now we can split $\lim ab=\lim a \cdot \lim b$, since both limits exists. Alternatively one can use L'ôpitals rule twice. $\lim_{x\to a} f/g = \lim_{x\to a} f'/g'$. Somewhat of a hazzle but it works

$$ \lim_{x\to 0} \frac{x^2}{\sin x \cdot \tan x} = \lim_{x\to 0} \frac{2x}{\cos x \tan x + \sin x \sec^2x} = \lim_{x\to 0} \frac{2}{\cos x + 1/\cos x + 2 \sin^2x \sec^2x} $$

The rest now follows from $\sin 0 = 0$ and $\cos 0 = 1$.

Answer 2

Note that the function under consideration is even and hence it is sufficient to consider $x \to 0^{+}$. Now we need to compare $x^{2}$ and $\sin x \tan x$ for $x > 0$. Let $$f(x) = x^{2} - \sin x\tan x$$ Then $f(0) = 0$ and $$f'(x) = 2x - \sin x\sec^{2}x - \sin x$$ Then we have $f'(0) = 0$. Further $$f''(x) = 2 - \sec x - 2\tan^{2}x - \cos x$$ and $f''(0) = 0$. Next we have $$f'''(x) = -\sec x\tan x - 4\tan x\sec^{2}x + \sin x = \sin x(1 - \sec^{2}x - 4\sec^{3}x)$$ As $x \to 0^{+}$ we can clearly see that $f'''(x) < 0$. This means that $f''(x)$ is decreasing and considering that $f''(0) = 0$ we have $f''(x) < 0$ when $x \to 0^{+}$. Continuing further in the same manner we see that $f(x) < 0$ when $x \to 0^{+}$. It follows that $\sin x\tan x > x^{2} > 0$ when $x \to 0^{+}$ so that $$0 < \dfrac{x^{2}}{\sin x\tan x} < 1$$ when $x \to 0^{+}$. Hence we have $$\left\lfloor\dfrac{x^{2}}{\sin x\tan x}\right\rfloor = 0$$ when $x \to 0^{+}$. Now it is obvious that the desired limit is $0$.

Answer 3

Let me continue where you stopped in your post.

Expand the denominator for a few terms and perform the long division. You should arrive to $$\frac{x^2}{\sin x\cdot \tan x} =1-\frac{x^2}{6}-\frac{7 x^4}{120}+O\left(x^5\right)$$ which is definitely smaller than $1$.

I am sure that you can take from here.