Question Number 128607 by john_santu last updated on 08/Jan/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}.\mathrm{cot}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$
Answered by malwan last updated on 08/Jan/21
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{x}}{{tan}\:{x}}\:−\:\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−{tan}\:{x}}{{x}^{\mathrm{2}} \:{tan}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−\left({x}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\:+…..\right)}{{x}^{\mathrm{2}} \:{tan}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\:\frac{−\frac{{x}}{\mathrm{3}}}{{tan}\:{x}}\:\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by liberty last updated on 08/Jan/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{xcos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$