Question Number 132537 by SLVR last updated on 15/Feb/21
$${If}\:{f}\left({x}\right)=\mathrm{8}{x}^{\mathrm{3}\:} +\mathrm{3}{x}\:{then}\:{lim}_{{x}\rightarrow\infty} \frac{{x}^{\mathrm{1}/\mathrm{3}} }{{f}^{−\mathrm{1}} \left(\mathrm{8}{x}\right)−{f}^{−\mathrm{1}} \left({x}\right)}\:{is} \\ $$
Commented by SLVR last updated on 15/Feb/21
$${can}\:{any}\:{one}\:{help}\:{me}...{please} \\ $$
Answered by Ñï= last updated on 16/Feb/21
$${x}=\mathrm{8}{t}^{\mathrm{3}} +\mathrm{3}{t} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{f}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{1}/\mathrm{3}} }=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\frac{{f}^{−\mathrm{1}} \left(\mathrm{8}{t}^{\mathrm{3}} +\mathrm{3}{t}\right)}{\left(\mathrm{8}{t}^{\mathrm{3}} +\mathrm{3}{t}\right)^{\mathrm{1}/\mathrm{3}} }=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\frac{{t}}{\left(\mathrm{8}{t}^{\mathrm{3}} +\mathrm{3}{t}\right)^{\mathrm{1}/\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{f}^{−\mathrm{1}} \left(\mathrm{8}{x}\right)}{{x}^{\mathrm{1}/\mathrm{3}} }=\mathrm{2}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{f}^{−\mathrm{1}} \left(\mathrm{8}{x}\right)}{\left(\mathrm{8}{x}\right)^{\mathrm{1}/\mathrm{3}} }=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{1}/\mathrm{3}} }{{f}^{−\mathrm{1}} \left(\mathrm{8}{x}\right)−{f}^{−\mathrm{1}} \left({x}\right)}=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2} \\ $$