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Question Number 130925 by Chhing last updated on 30/Jan/21

$$$$$$\mathrm{Problem}$$$$\mathrm{Without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\mathrm{calculate}$$$$\underset{x\to 0}{\lim }\frac{{\tan }^2\left(\mathrm{x}\right)-{\mathrm{x}}^2\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2-{\sin }^2\left(\mathrm{x}\right)}$$

Commented by malwan last updated on 30/Jan/21

$$x\to 0$$$$\Rightarrow tanx=sinx\wedge cos2x=1$$$$\underset{x\to 0}{lim}\frac{{sin}^{2}x-x^2}{x^2-{sin}^{2}x}=-1$$$$Itisnotright$$$$whay?$$

Commented by Chhing last updated on 30/Jan/21

$$\mathrm{but}\tan \left(0\right)=0$$

Commented by malwan last updated on 30/Jan/21

$$andsin\left(0\right)=0$$

Commented by Chhing last updated on 31/Jan/21

$$\mathrm{Yes},\mathrm{sir}$$

Answered by bramlexs22 last updated on 30/Jan/21

$$=\underset{x\to 0}{\lim }\frac{{(x+\frac{x^3}{3})}^2-x^2(1-2x^2)}{x^2-{(x-\frac{x^3}{6})}^2}$$$$=\underset{x\to 0}{\lim }\frac{x^2{(1+\frac{x^2}{3})}^2-x^2(1-2x^2)}{x^2(1-{(1-\frac{x^2}{6})}^2)}$$$$=\underset{x\to 0}{\lim }\frac{(1+\frac{2x^2}{3})-1+2x^2}{1-(1-\frac{x^2}{3})}$$$$=\underset{x\to 0}{\lim }\frac{\left(\frac{8x^2}{3}\right)}{\left(\frac{x^2}{3}\right)}$$$$=8✓$$

Commented by bramlexs22 last updated on 30/Jan/21

Answered by benjo_mathlover last updated on 30/Jan/21

$$\underset{x\to 0}{\lim }\frac{\tan {}^2\mathrm{x}-{\mathrm{x}}^2(\cos {}^2\mathrm{x}-\sin {}^2\mathrm{x})}{{\mathrm{x}}^2-\sin {}^2\mathrm{x}}$$$$\underset{x\to 0}{\lim }\frac{\tan {}^2\mathrm{x}-{\mathrm{x}}^2\cos {}^2\mathrm{x}+{\mathrm{x}}^2\sin {}^2\mathrm{x}}{(\mathrm{x}+\sin \mathrm{x})(\mathrm{x}-\sin \mathrm{x})}$$$$\underset{x\to 0}{\lim }\frac{(\tan \mathrm{x}+\mathrm{xcos}\mathrm{x})(\tan \mathrm{x}-\mathrm{xcos}\mathrm{x})+{\mathrm{x}}^2\sin {}^2\mathrm{x}}{(\mathrm{x}+\sin \mathrm{x})(\mathrm{x}-\sin \mathrm{x})}$$$$\underset{x\to 0}{\lim }\frac{(\frac{\tan \mathrm{x}}{\mathrm{x}}+\cos \mathrm{x})(\mathrm{x}+\frac{{\mathrm{x}}^3}{3}-\mathrm{x}(1-\frac{1}{2}{\mathrm{x}}^2))+{\mathrm{x}}^2(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})}{(1+\frac{\sin \mathrm{x}}{\mathrm{x}})(\mathrm{x}-\mathrm{x}+\frac{{\mathrm{x}}^3}{6})}$$$$\underset{x\to 0}{\lim }\frac{2(\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^3}{2})+{\mathrm{x}}^3-\frac{{\mathrm{x}}^5}{6}}{2\left(\frac{{\mathrm{x}}^3}{6}\right)}=\underset{x\to 0}{\lim }\frac{\frac{8}{3}-\frac{{\mathrm{x}}^2}{6}}{\frac{1}{3}}$$$$=8.$$

Commented by bramlexs22 last updated on 30/Jan/21

$$yeahhh...✓$$

Commented by Chhing last updated on 30/Jan/21

$$\mathrm{Thank}\mathrm{you}$$

Commented by Chhing last updated on 30/Jan/21

$$\mathrm{Thank}\mathrm{you}$$

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