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Question Number 197282 by cortano12 last updated on 12/Sep/23

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

Answered by MM42 last updated on 12/Sep/23

$${lim}_{x\to 0}\frac{{sin}^{2}x-x^2+x^2-{sinx}^2}{x^2({cos}^{2}x-1+1-{cosx}^2)}$$$$={lim}_{x\to 0}\frac{(sinx+x)(sinx-x)+\frac{1}{6}{x}^6}{x^2((cosx-1)(cosx+1)+\frac{1}{2}{x}^4)}$$$$={lim}_{x\to 0}\frac{\left(2x\right)(-\frac{1}{6}{x}^3)+\frac{1}{6}{x}^6}{x^2((-\frac{1}{2}{x}^2)(cosx+1)+\frac{1}{2}{x}^4)}$$$$={lim}_{x\to 0}\frac{-\frac{1}{3}{x}^4+\frac{1}{6}{x}^6}{-x^4+\frac{1}{2}{x}^6}=\frac{1}{3}✓$$$$$$

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