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solve-lim-x-0-x-2-tan-sinpix-2x-solution-let-L-lim-x-0-x-2-tan-sinpix-2x-since-sinx-x-x-3-6-L-lim-x-0-x-2-tan-pix-2x-pi-3-x-3-12x-L-lim-x-0-x-2-tan-pi-2-pi-3-x-2




Question Number 192573 by senestro last updated on 21/May/23
solve;  lim_(x→0) x^2 tan(((sinπx)/(2x)))  solution  let L=lim_(x→0) x^2 tan(((sinπx)/(2x)))  since sinx∼x−(x^3 /6)   L=lim_(x→0) x^2 tan(((πx)/(2x))−((π^3 x^3 )/(12x)))  L=lim_(x→0) x^2 tan((π/2)−((π^3 x^2 )/(12)))  since tan((π/2)−x)=(1/(tanx))  L=lim_(x→0) (x^2 /(tan(((π^3 x^2 )/(12)))))  L=lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12))))) ((12)/π^3 )  L=((12)/π^3 )lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12)))))  since lim_(x→0) (x/(tanx))=1  L=((12)/π^3 ) ∙1=((12)/π^3 )  solved by HY a.k.a senestro
$${solve}; \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{{sin}\pi{x}}{\mathrm{2}{x}}\right) \\ $$$${solution} \\ $$$${let}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{{sin}\pi{x}}{\mathrm{2}{x}}\right) \\ $$$${since}\:{sinx}\sim{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\: \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{\pi{x}}{\mathrm{2}{x}}−\frac{\pi^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{12}{x}}\right) \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$${since}\:{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{1}}{{tanx}} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} }{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}}{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)}\:\frac{\mathrm{12}}{\pi^{\mathrm{3}} } \\ $$$${L}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} }\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}}{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)} \\ $$$${since}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{{tanx}}=\mathrm{1} \\ $$$${L}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} }\:\centerdot\mathrm{1}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} } \\ $$$${solved}\:{by}\:{HY}\:{a}.{k}.{a}\:{senestro} \\ $$

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