Question Number 115162 by john santu last updated on 24/Sep/20 | ||
$$\:\left(\mathrm{1}\right)\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right)\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}^{\mathrm{3}} +\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \:=? \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{csc}\:{x}\:.\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:=? \\ $$ | ||
Answered by bobhans last updated on 24/Sep/20 | ||
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right)\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{{x}}}{\left(\mathrm{1}−\sqrt{{x}}\right)\left(\mathrm{1}+\sqrt{{x}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{{x}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{6}} \\ $$ | ||
Answered by bobhans last updated on 24/Sep/20 | ||
$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}^{\mathrm{3}} \:+\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \:=\:{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} +\sqrt{\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \\ $$$$\:=\:{e}^{\infty} \:=\:\infty\: \\ $$ | ||
Answered by bobhans last updated on 24/Sep/20 | ||
$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{csc}\:{x}.\:\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}} \\ $$$$\left[\:{let}\:\mathrm{sin}\:{x}\:=\:{w}\:;\:{w}\rightarrow\mathrm{0}\:\right] \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}}\:=\:\underset{{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({w}\right)}{{w}}\:=\:\mathrm{1} \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 24/Sep/20 | ||
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\mathrm{1}+\mathrm{x}}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{x}}}\right)=\frac{\pi}{\mathrm{6}} \\ $$ | ||