Question and Answers Forum

Limits Questions

Question Number 101422 by john santu last updated on 02/Jul/20

$$\underset{\mathrm{h}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\left(\alpha+\mathrm{h}\right)^{\mathrm{2}} \right)−\mathrm{sin}\:\left(\alpha^{\mathrm{2}} \right)}{\mathrm{cos}\:\left(\left(\alpha+\mathrm{h}\right)^{\mathrm{2}} \mathrm{sin}\:\left(\alpha+\mathrm{h}\right)−\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{sin}\:\left(\alpha\right)\right.}\:=? \\$$

Commented by john santu last updated on 02/Jul/20

$$\Leftrightarrow\frac{\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\left(\alpha+\mathrm{h}\right)^{\mathrm{2}} −\mathrm{sin}\:\left(\alpha^{\mathrm{2}} \right)\right.}{\mathrm{h}}}{\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\left(\alpha+\mathrm{h}\right)^{\mathrm{2}} \mathrm{sin}\:\left(\alpha+\mathrm{h}\right)−\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{sin}\:\left(\alpha\right)\right.}{\mathrm{h}}}\:= \\$$$$\mathrm{let}\:\mathrm{f}\left(\alpha\right)\:=\:\mathrm{sin}\:\left(\alpha^{\mathrm{2}} \right)\:\rightarrow\mathrm{f}\:'\left(\alpha\right)=\:\mathrm{2}\alpha\:\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right) \\$$$$\mathrm{let}\:\mathrm{g}\left(\alpha\right)\:=\:\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{sin}\:\left(\alpha\right)\rightarrow\mathrm{g}\:'\left(\alpha\right)=−\mathrm{2}\alpha\mathrm{sin}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{sin}\:\left(\alpha\right)+\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{cos}\:\left(\alpha\right) \\$$$$\mathrm{therefore}\:\mathrm{we}\:\mathrm{get}\:\mathrm{limit} \\$$$$=\:\frac{\mathrm{2}\alpha\:\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)}{−\mathrm{2}\alpha\:\mathrm{sin}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{sin}\:\left(\alpha\right)+\mathrm{cos}\:\left(\alpha^{\mathrm{2}} \right)\mathrm{cos}\:\left(\alpha\right)}\:\bigstar \\$$

Commented by Dwaipayan Shikari last updated on 02/Jul/20

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left(\left(\alpha+{h}\right)^{\mathrm{2}} \right)−{sin}\left(\alpha^{\mathrm{2}} \right)}{{h}}.\frac{{h}}{{cos}\left(\left(\alpha+{h}\right)^{\mathrm{2}} \right){sin}\left(\alpha+{h}\right)−{cos}\alpha^{\mathrm{2}} {sin}\alpha} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}\alpha{cos}\alpha^{\mathrm{2}} }{−\mathrm{2}\alpha{sin}\alpha^{\mathrm{2}} {sin}\alpha+{cos}\alpha{cos}\alpha^{\mathrm{2}} }=\frac{\mathrm{2}\alpha}{−\mathrm{2}\alpha{tan}\alpha^{\mathrm{2}} {sin}\alpha+{cos}\alpha}\bigstar\blacksquare\mathscr{L} \\$$