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Alright this looks like a very simple problem at the first go.

I need to find $$\lim_{x\rightarrow0^+} { \left\lfloor{\frac{x}{\sin (x)}}\right\rfloor}$$

So since I know the inner function's graph beforehand I know the answer will be 1.

But now here's my problem.When I'm trying to find which function $x$ or $\sin(x)$ is greater when $x \rightarrow 0^+$ I take a function $g(x)=x-\sin(x)$ and find its derivative.So $g'(x)=1-\cos(x)$.Now when I find $g'(x \rightarrow 0^+)$ I get $0$ !

So I'm not being able to prove that $g(x)$ is increasing when $x$ is slightly greater than $0$.And thus I can't prove that $x>sin(x)$ when $x$ is slighty greater than $0$ and neither can I prove $\frac{x}{\sin (x)}>1$.Where am I going wrong?

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  • $\begingroup$ I think you're thinking of $\displaystyle{\sin x \over x}$, TheGreatDuck. $\endgroup$ Jan 29, 2016 at 6:00

2 Answers 2

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For $g(x) = x - \sin x$ we have $g'(x) = 1 - \cos x > 0$ for $0 < x \leqslant \pi/2$. and $g(0) = 0$. By the mean value theorem for any $x$ in the interval there exists $\xi$ between $0$ and $x$ such that $g(x) = g'(\xi)x > 0.$

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  • $\begingroup$ What does this $g(x) = g'(\xi)x > 0$ mean ? $\endgroup$
    – user220382
    Jan 26, 2016 at 6:05
  • $\begingroup$ Can you please clarify that last notation? $\endgroup$
    – user220382
    Jan 26, 2016 at 6:06
  • $\begingroup$ I added above. Use the mean value theorem $g(x) - g(0) = g'(\xi)(x- 0)$ for some $\xi$ between $0$ and $x$. $\endgroup$
    – RRL
    Jan 26, 2016 at 6:08
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There is a "classical" proof (based on geometry and trigonometry) that $\displaystyle\lim_{x\to 0} {\sin x \over x} = 1$. In it, it is proven that $\sin x < x < \tan x$ when $x>0$ and is small. It should be in any Calculus textbook.

Dividing this equation through by $\sin x$, you get $\displaystyle 1 < {x\over \sin x}$, when $x>0$ and $x$ is small. Hence $\displaystyle\left\lfloor {x\over \sin x}\right\rfloor \ge 1$ for those same $x$.

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  • $\begingroup$ Read the question.... $\endgroup$
    – user220382
    Jan 26, 2016 at 5:56
  • $\begingroup$ There might be several ways of proving that claim.But what I am looking for is a calculus based proof (without taylor series).And my doubt is why is the general method of taking the difference of two functions and finding its derivative to find out which function is greater is not working here. $\endgroup$
    – user220382
    Jan 26, 2016 at 5:59
  • $\begingroup$ Looking at your original post, $g'(x)>0$ when $x>0$ and small, because $\cos x < 1$. Thus, $g$ is increasing on $(0,\varepsilon)$ for small $\epsilon$. (And, BTW, I didn't use Taylor Series.) $\endgroup$ Jan 26, 2016 at 6:04
  • $\begingroup$ Okay thanks :-) $\endgroup$
    – user220382
    Jan 26, 2016 at 6:09
  • $\begingroup$ I think you're thinking of $\displaystyle{\sin x \over x}$, TheGreatDuck. $\endgroup$ Jan 29, 2016 at 5:59

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