Question Number 176638 by cortano1 last updated on 23/Sep/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)\right)}=? \\ $$
Answered by a.lgnaoui last updated on 23/Sep/22
![((sin (x^2 )[((sin^2 (x))/(sin (x^2 )))−1])/(x^2 cos(x^2 )[((cos^2 (x))/(cos (x^2 )))−1]))=((tan (x^2 ))/x^2 )×((sin^2 (x)[(1/(sin (x^2 )))−(1/(sin^2 (x)))])/(cos^2 (x)[(1/(cos (x^2 )))−(1/(cos^2 (x)))])) =((tan (x^2 )tan^2 (x))/x^2 )(([((1/(sin (x^2 )))−1−cot^2 (x))])/([(1/(cos (x^2 )))−1−tan^2 (x))])) =((tan (x^2 )tan^2 (x)((1/(sin (x^2 )))−1)−tan (x^2 ))/(x^2 [(((1−cos (x^2 ))/(cos (x^2 )))−tan^2 (x))])) =((tan (x^2 )tan^2 (x)((1/(sin (x^2 )))−1)−tan (x^2 ))/(x^2 [((1/(cos (x^2 )))−1)−tan^2 (x)])) =((tan (x^2 ))/x^2 ) ×((tan^2 (x)[(1/(sin (x^2 )))−2])/([(1/(cos (x^2 )))−1−tan^2 (x)] )) (√(1/(cos (x^2 )))) =(√(1+tan^2 (x^2 ))) lim_(x→0) ((sin^2 (x)−sin (x^2 ))/(x^2 cos^2 (x)−cos(x^2 ))) = lim_(x→0) ((tan^2 (x)[(1/(sin (x^2 )))−2])/( [(√(1+tan^2 (x^2 ))) −1−tan^2 (x) )) =1×lim_(x→0) [−((tan^2 (x)((1/(sin (x^2 )))−2) )/(tan^2 (x)))] =−2+lim_(x→0) (1/(sin (x^2 )))=+∞](https://www.tinkutara.com/question/Q176650.png)
$$\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)\left[\frac{\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}\right]}{{x}^{\mathrm{2}} \mathrm{cos}\left({x}^{\mathrm{2}} \right)\left[\frac{\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}\right]}=\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }×\frac{\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)\left[\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)}\right]}{\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)\left[\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)}\right]} \\ $$$$=\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }\frac{\left[\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} \left({x}\right)\right)\right]}{\left.\left[\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\right)\right]} \\ $$$$=\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}\right)−\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \left[\left(\frac{\mathrm{1}−\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{tan}^{\mathrm{2}} \:\left({x}\right)\right)\right]} \\ $$$$=\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}\right)−\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \left[\left(\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}\right)−\mathrm{tan}^{\mathrm{2}} \:\left({x}\right)\right]} \\ $$$$=\frac{\mathrm{tan}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:×\frac{\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\left[\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{2}\right]}{\left[\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left({x}\right)\right]\:} \\ $$$$\:\:\sqrt{\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}^{\mathrm{2}} \right)}}\:\:=\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}\:\:\: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)−\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \left({x}\right)−\mathrm{cos}\left({x}^{\mathrm{2}} \right)}\:=\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{tan}^{\mathrm{2}} \left({x}\right)\left[\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{2}\right]}{\:\left[\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}\:−\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left({x}\right)\:\right.} \\ $$$$=\mathrm{1}×{lim}_{{x}\rightarrow\mathrm{0}} \left[−\frac{\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}−\mathrm{2}\right)\:}{\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)}\right]\:=−\mathrm{2}+{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}=+\infty \\ $$$$ \\ $$
Commented by cortano1 last updated on 24/Sep/22
