Question Number 210674 by noraell last updated on 15/Aug/24
Answered by BHOOPENDRA last updated on 16/Aug/24
$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} \:\left\{−\mathrm{1}+\sqrt{\left.−\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right\}}\:\right.} \\ $$$$\left\{{lim}\:{x}\rightarrow\mathrm{0}\:\frac{{sinx}}{{x}}=\mathrm{1},{limx}\rightarrow\mathrm{0}\:\frac{{tanx}}{{x}}=\mathrm{1}\right\} \\ $$$${conjugate}\:{multiplication} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\frac{−\mathrm{1}−\sqrt{\left(−\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{4}{k}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow−{limx}\rightarrow\mathrm{0}\:\:\frac{−\mathrm{1}−\sqrt{\left(−\mathrm{4}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{1}\right)}}{\mathrm{4}{k}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\:{where}\:{k}\neq\mathrm{0} \\ $$