Question Number 188575 by mathlove last updated on 03/Mar/23
Answered by universe last updated on 03/Mar/23
$$\sqrt{{e}} \\ $$
Answered by floor(10²Eta[1]) last updated on 03/Mar/23
$$\mathrm{L}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}e}^{\left(\frac{\mathrm{5x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5}}−\mathrm{2}\right)} \\ $$$$\mathrm{lnL}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{5x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5}}−\mathrm{2}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{12}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{L}=\sqrt{\mathrm{e}} \\ $$
Answered by manxsol last updated on 03/Mar/23
$${lim}^{} \underset{{x}\rightarrow−\infty} {\:}{e}^{\left(\frac{\frac{\mathrm{5}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}{\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}−\mathrm{2}\right)} \\ $$$${lim}_{{x}\rightarrow−\infty} {e}^{\left(\frac{\mathrm{5}−\frac{\mathrm{2}}{{x}^{\mathrm{2}\:} }}{\mathrm{2}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}−\mathrm{2}\right)} \\ $$$${lim}_{{x}\rightarrow−\infty} {e}^{\left(\frac{\mathrm{5}−\mathrm{0}}{\mathrm{2}−\mathrm{0}}−\mathrm{2}\right)} \\ $$$${lim}_{{x}\rightarrow−\infty} {e}^{\mathrm{0}.\mathrm{5}} =\sqrt{{e}}\:\: \\ $$
Answered by CElcedricjunior last updated on 03/Mar/23
$$\sqrt{\boldsymbol{{e}}} \\ $$