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Question Number 85061 by M±th+et£s last updated on 18/Mar/20

$$\underset{x\to 0}{lim}\frac{xtan2x-2xtan\left(x\right)}{{(1-cos\left(2x\right))}^2}$$

Commented by mathmax by abdo last updated on 18/Mar/20

$$letf\left(x\right)=\frac{xtan\left(2x\right)-2xtan\left(x\right)}{{(1-cos\left(2x\right))}^2}wehave$$$$tanx=x+\frac{x^3}{3}+o\left(x^3\right)andtan\left(2x\right)=2x+\frac{8}{3}{x}^3+o\left(x^3\right)$$$$\Rightarrow tanx\sim x+\frac{x^3}{3}andtan\left(2x\right)\sim 2x+\frac{8}{3}{x}^{3}and$$$$1-cos\left(2x\right)\sim \frac{{\left(2x\right)}^2}{2}=2x^2\Rightarrow f\left(x\right)\sim \frac{2x^2+\frac{8}{3}{x}^4-2x^2-\frac{2}{3}{x}^4}{4x^4}$$$$\Rightarrow f\left(x\right)\sim \frac{2x^4}{4x^4}=\frac{1}{2}\Rightarrow {lim}_{x\to 0}f\left(x\right)=\frac{1}{2}$$

Commented by M±th+et£s last updated on 18/Mar/20

$$thanks$$

Commented by mathmax by abdo last updated on 18/Mar/20

$$youarewelcome$$

Answered by john santu last updated on 19/Mar/20

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}\tan 2\mathrm{x}-2\mathrm{x}\tan \mathrm{x}}{{(1-1+2\sin {}^2\mathrm{x})}^2}=$$$$\frac{1}{4}\underset{x\to 0}{\lim }\frac{\mathrm{x}(\tan 2\mathrm{x}-2\tan \mathrm{x})}{\sin {}^4\mathrm{x}}=$$$$\frac{1}{4}\underset{x\to 0}{\lim }\frac{2\tan \mathrm{x}(\frac{1}{1-\tan {}^2\mathrm{x}}-1)}{\sin {}^3\mathrm{x}}=$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{\tan {}^2\mathrm{x}}{\sin {}^2\mathrm{x}}\times \underset{x\to 0}{\lim }\frac{1}{1-\tan {}^2\mathrm{x}}=$$$$\frac{1}{2}\therefore$$

Commented by M±th+et£s last updated on 19/Mar/20

$$thankyousir$$

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