Question Number 213073 by ajfour last updated on 29/Oct/24
$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}−\mathrm{1}\right)\right\} \\ $$
Answered by universe last updated on 29/Oct/24
$${L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${L}\:=\:\frac{\mathrm{1}}{{a}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }−{a}}{{x}^{\mathrm{2}} } \\ $$$${L}\:=\frac{\mathrm{1}}{{a}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}/\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{\mathrm{2}{x}}\:\: \\ $$$${L}\:=\:\mathrm{1}/\mathrm{2}{a}^{\mathrm{2}} \:\: \\ $$$$ \\ $$