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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?

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First, what can YOU say about it? –  Did Jun 11 at 9:19
    
:) good comment ... well let me try .. :) –  Debashish Jun 11 at 9:22
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It might help to note that $|\sin x|<|x|$ for $x\ne0$. –  Gerry Myerson Jun 11 at 9:39
    
@ Gerry Myerson ... Thanks.. I think I have the answer now. Please check if its correct. I have edited it in my question as there is no "Answer" option, my post being on hold. –  Debashish Jun 11 at 12:06
    
@ Did and others .. I think the hold on my question may be removed .. if you think its ok now. –  Debashish Jun 11 at 12:12

1 Answer 1

up vote 1 down vote accepted

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 $198$.

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