Question Number 194286 by mokys last updated on 02/Jul/23
$$ \\ $$$$\:\:\boldsymbol{{find}}\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\lfloor\:\frac{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\rfloor \\ $$
Answered by tri26112004 last updated on 02/Jul/23
$$=\:\mathrm{1} \\ $$
Answered by Skabetix last updated on 02/Jul/23
$$\frac{{tan}\left({x}\right)}{{x}}=\frac{{sin}\left({x}\right)}{{x}×{cos}\left({x}\right)}=\frac{{sin}\left({x}\right)}{{x}}×\frac{\mathrm{1}}{{cos}\left({x}\right)} \\ $$$${Lim}_{{x}\rightarrow\mathrm{0}} \frac{{tan}\left({x}\right)}{{x}}={Lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sin}\left({x}\right)}{{x}}×\frac{\mathrm{1}}{{cos}\left({x}\right)}\right) \\ $$$$=\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$
Commented by mokys last updated on 02/Jul/23
$${sory}\:{sir}\:{but}\:{this}\:{floor}\:{function}\:\lfloor\:\frac{{tan}\left({x}\right)}{{x}}\rfloor\:\neq\:\frac{{tan}\left({x}\right)}{{x}}\: \\ $$$${and}\:{the}\:{answer}\:{is}\:\mathrm{2} \\ $$
Commented by Frix last updated on 02/Jul/23
![((d[tan x])/dx)=(1/(cos^2 x))>1∀x∈(0, 1)∪(−1,0) ⇒ 1<((tan x)/x)<1.56∀x∈(0, 1)∪(−1,0) ⇒ ⌊((tan x)/x)⌋=1∀x∈(0, 1)∪(−1,0)](https://www.tinkutara.com/question/Q194294.png)
$$\frac{{d}\left[\mathrm{tan}\:{x}\right]}{{dx}}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}>\mathrm{1}\forall{x}\in\left(\mathrm{0},\:\mathrm{1}\right)\cup\left(−\mathrm{1},\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{1}<\frac{\mathrm{tan}\:{x}}{{x}}<\mathrm{1}.\mathrm{56}\forall{x}\in\left(\mathrm{0},\:\mathrm{1}\right)\cup\left(−\mathrm{1},\mathrm{0}\right)\:\Rightarrow \\ $$$$\lfloor\frac{\mathrm{tan}\:{x}}{{x}}\rfloor=\mathrm{1}\forall{x}\in\left(\mathrm{0},\:\mathrm{1}\right)\cup\left(−\mathrm{1},\mathrm{0}\right) \\ $$