Question Number 108235 by mathdave last updated on 15/Aug/20
Answered by Dwaipayan Shikari last updated on 15/Aug/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{3}{x}\right)}{{sin}^{−\mathrm{1}} \mathrm{2}{x}}=\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{2}{x}}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\: \\ $$$${When}\:{x}\rightarrow\mathrm{0} \\ $$$${tan}^{−\mathrm{1}} {x}\rightarrow{x} \\ $$$${sin}^{−\mathrm{1}} {x}\rightarrow{x} \\ $$