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Question Number 78596 by jagoll last updated on 19/Jan/20

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

Commented by mathmax by abdo last updated on 19/Jan/20

$$letA\left(x\right)={(1-3{tan}^{2}x)}^{\frac{2}{{sin}^2\left(3x\right)}}\Rightarrow A\left(x\right)=e^{\frac{2}{{sin}^2\left(3x\right)}ln(1-3{tan}^{2}x)}$$$$wehavetanx\sim x\Rightarrow 1-3{tan}^{2}x\sim 1-3x^{2}andln(1-3x^2)\sim -3x^2$$$$alsosin\left(3x\right)\sim 3x\Rightarrow {sin}^2\left(3x\right)\sim 9x^2\Rightarrow$$$$\frac{2}{{sin}^2\left(3x\right)}ln(1-3{tan}^{2}x)\sim \frac{-6x^2}{9x^2}=-\frac{2}{3}\Rightarrow {lim}_{x\to 0}A\left(x\right)=e^{-\frac{2}{3}}$$$$=\frac{1}{({}^3\sqrt{e^2)}}$$

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

$$weknowthat\underset{x\to 0}{\lim }{(1+x)}^{\frac{1}{x}}=e$$$$nowweworkthequestion$$$$\underset{x\to 0}{\lim }\left[{(1+(-3\tan {}^{2}x))}^{\frac{1}{-3\tan {}^{2}x}}\right]{}^{\frac{-6\tan {}^{2}x}{\sin {}^{2}3x}}$$$$=e{}^{\underset{x\to 0}{\lim }\left(\frac{-6\tan {}^{2}x}{\sin {}^{2}3x}\right)}=e{}^{-\frac{6}{9}}=\frac{1}{\sqrt[3]{e^2}}.$$

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