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Question Number 194286 by mokys last updated on 02/Jul/23

$$$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{d}\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\lfloor \frac{\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)}{\mathit{x}}\rfloor$$

Answered by tri26112004 last updated on 02/Jul/23

$$=1$$

Answered by Skabetix last updated on 02/Jul/23

$$\frac{tan\left(x\right)}{x}=\frac{sin\left(x\right)}{x\times cos\left(x\right)}=\frac{sin\left(x\right)}{x}\times \frac{1}{cos\left(x\right)}$$$${Lim}_{x\to 0}\frac{tan\left(x\right)}{x}={Lim}_{x\to 0}\left(\frac{sin\left(x\right)}{x}\times \frac{1}{cos\left(x\right)}\right)$$$$=1\times 1=1$$

Commented by mokys last updated on 02/Jul/23

$$sorysirbutthisfloorfunction\lfloor \frac{tan\left(x\right)}{x}\rfloor \ne \frac{tan\left(x\right)}{x}$$$$andtheansweris2$$

Commented by Frix last updated on 02/Jul/23

$$\frac{d\left[\tan x\right]}{dx}=\frac{1}{{\cos }^{2}x}>1\mathrm{\forall }x\in (0,1)\cup (-1,0)\Rightarrow$$$$1<\frac{\tan x}{x}<1.56\mathrm{\forall }x\in (0,1)\cup (-1,0)\Rightarrow$$$$\lfloor \frac{\tan x}{x}\rfloor =1\mathrm{\forall }x\in (0,1)\cup (-1,0)$$

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