Question Number 74870 by vishalbhardwaj last updated on 02/Dec/19
![solve with explanation lim_(x→0^− ) [(x/(sinx))], where [ ] represents greatest integer](https://www.tinkutara.com/question/Q74870.png)
$$\mathrm{solve}\:\mathrm{with}\:\mathrm{explanation} \\ $$$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}^{−} } {\mathrm{m}}\left[\frac{\mathrm{x}}{\mathrm{sinx}}\right],\:\mathrm{where}\:\left[\:\:\right]\:\mathrm{represents}\:\mathrm{greatest}\:\mathrm{integer} \\ $$
Commented by mathmax by abdo last updated on 02/Dec/19
![we have sinx =Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!)) with radius R=+∞ sinx =x−(x^3 /(3!)) +(x^5 /(5!))−.... ⇒ x−(x^3 /6) ≤sinx ≤x (we can take 0≤x≤(π/2)) 1−(x^2 /6)≤((sinx)/x) ≤1 ⇒[1−(x^2 /6)] ≤[((sinx)/x)]≤1 ⇒1+[−(x^2 /2)]≤[((sinx)/x)]≤1 ⇒lim_(x→0) [((sinx)/x)] =1](https://www.tinkutara.com/question/Q74873.png)
$${we}\:{have}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:{with}\:{radius}\:{R}=+\infty \\ $$$${sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−….\:\Rightarrow\:\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant{sinx}\:\leqslant{x}\:\left({we}\:{can}\:{take}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\leqslant\frac{{sinx}}{{x}}\:\leqslant\mathrm{1}\:\Rightarrow\left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right]\:\leqslant\left[\frac{{sinx}}{{x}}\right]\leqslant\mathrm{1}\:\Rightarrow\mathrm{1}+\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]\leqslant\left[\frac{{sinx}}{{x}}\right]\leqslant\mathrm{1} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\left[\frac{{sinx}}{{x}}\right]\:=\mathrm{1} \\ $$
Answered by mind is power last updated on 02/Dec/19
![(x/(sin(x)))>0 whe x∈]−(π/2),0[ [(x/(sin(x)))] ≤(x/(sin(x)))...(√E) x<sin(x), for all x∈]−(π/2),0[ proof cos(t)≤1 f(x)=x−sin(x)⇒f′(x)=1−cos(x)≥0 f increase f(0)=0⇒ x−sin(x)<0 ∀x∈[−(π/2),0[ ⇒x<sin(x) ⇒(x/(sin(x)))≥1 , ∴sin(x)≤0∴ ⇒[(x/(sin(x)))]≥1 ⇒E⇔1≤[(x/(sin(x)))]≤(x/(sin(x))) lim_(x→0) (x/(sin(x)))=lim_(x→0) (1/(cos(x)))=1 hopitaks Rulls ⇒lim_(x→0) [(x/(sin(x)))]=1](https://www.tinkutara.com/question/Q74872.png)
$$\left.\:\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}>\mathrm{0}\:\mathrm{whe}\:\mathrm{x}\in\right]−\frac{\pi}{\mathrm{2}},\mathrm{0}\left[\right. \\ $$$$\:\:\:\:\:\:\left[\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}\right]\:\:\:\leqslant\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}…\sqrt{\mathrm{E}} \\ $$$$\left.\mathrm{x}<\mathrm{sin}\left(\mathrm{x}\right),\:\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\right]−\frac{\pi}{\mathrm{2}},\mathrm{0}\left[\right. \\ $$$$\mathrm{proof} \\ $$$$\mathrm{cos}\left(\mathrm{t}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}−\mathrm{sin}\left(\mathrm{x}\right)\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{1}−\mathrm{cos}\left(\mathrm{x}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{f}\:\mathrm{increase}\:\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\:\:\:\:\mathrm{x}−\mathrm{sin}\left(\mathrm{x}\right)<\mathrm{0}\:\forall\mathrm{x}\in\left[−\frac{\pi}{\mathrm{2}},\mathrm{0}\left[\right.\right. \\ $$$$\Rightarrow\mathrm{x}<\mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}\geqslant\mathrm{1}\:\:\:,\:\:\therefore\mathrm{sin}\left(\mathrm{x}\right)\leqslant\mathrm{0}\therefore \\ $$$$\Rightarrow\left[\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}\right]\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{E}\Leftrightarrow\mathrm{1}\leqslant\left[\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}\right]\leqslant\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{x}\right)}=\mathrm{1}\:\:\mathrm{hopitaks}\:\mathrm{Rulls} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{x}\right)}\right]=\mathrm{1} \\ $$