What can you say about the following limit:
$$\lim_{x\rightarrow 0}\left(\left[\dfrac{100x}{\sin x}\right]+\left[\dfrac{99\sin x}{x}\right]\right)$$ where [.] represents the greatest integer function (floor function).
Well I see that my question is on hold. I have got some idea from Gerry Meyerson. Let me know if the following is acceptable:
Since $|\sin x|<|x|$ for $x\neq 0$, so $\dfrac{x}{\sin x}\rightarrow 1^+$ as $x\rightarrow 0$ and $\dfrac{\sin x}{x}\rightarrow 1^-$ as $x\rightarrow 0$.
Thus $\left[\dfrac{100x}{\sin x}\right]\rightarrow 100$ as $x\rightarrow 0$ and $\left[\dfrac{99\sin x}{x}\right]\rightarrow 98$ as $x\rightarrow 0$. So, I think the limit evaluates to $198$. Am I correct?
$\endgroup$ 51 Answer
$\begingroup$Well .. thanks to all the useful comments. I think I have the answer now. Still I would appreciate further suggestion.
Since $|\sin x|<|x|$ for $x\neq 0$, so $\dfrac{x}{\sin x}\rightarrow 1^+$ as $x\rightarrow 0$ and $\dfrac{\sin x}{x}\rightarrow 1^-$ as $x\rightarrow 0$.
Thus $\left[\dfrac{100x}{\sin x}\right]\rightarrow 100$ as $x\rightarrow 0$ and $\left[\dfrac{99\sin x}{x}\right]\rightarrow 98$ as $x\rightarrow 0$. So, I think the limit evaluates to $198$.
$\endgroup$
