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Question Number 192573 by senestro last updated on 21/May/23

$$solve;$$$$\underset{x\to 0}{\lim }{x}^{2}tan\left(\frac{sin\pi x}{2x}\right)$$$$solution$$$$letL=\underset{x\to 0}{\lim }{x}^{2}tan\left(\frac{sin\pi x}{2x}\right)$$$$sincesinx\sim x-\frac{x^3}{6}$$$$L=\underset{x\to 0}{\lim }{x}^{2}tan(\frac{\pi x}{2x}-\frac{{\pi }^3{x}^3}{12x})$$$$L=\underset{x\to 0}{\lim }{x}^{2}tan(\frac{\pi }{2}-\frac{{\pi }^3{x}^2}{12})$$$$sincetan(\frac{\pi }{2}-x)=\frac{1}{tanx}$$$$L=\underset{x\to 0}{\lim }\frac{x^2}{tan\left(\frac{{\pi }^3{x}^2}{12}\right)}$$$$L=\underset{x\to 0}{\lim }\frac{\frac{{\pi }^3{x}^2}{12}}{tan\left(\frac{{\pi }^3{x}^2}{12}\right)}\frac{12}{{\pi }^3}$$$$L=\frac{12}{{\pi }^3}\underset{x\to 0}{\lim }\frac{\frac{{\pi }^3{x}^2}{12}}{tan\left(\frac{{\pi }^3{x}^2}{12}\right)}$$$$since\underset{x\to 0}{\lim }\frac{x}{tanx}=1$$$$L=\frac{12}{{\pi }^3}\cdot 1=\frac{12}{{\pi }^3}$$$$solvedbyHYa.k.asenestro$$

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