Question Number 21179 by Joel577 last updated on 15/Sep/17
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{16}{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}}\:−\:\sqrt{{x}^{\mathrm{2}} }\:−\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}} \\ $$
Answered by dioph last updated on 15/Sep/17
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\sqrt{\mathrm{16}+\frac{\mathrm{4}}{{x}}}\:−\:{x}\:−\:{x}\sqrt{\mathrm{9}+\frac{\mathrm{3}}{{x}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{\mathrm{16}+\frac{\mathrm{4}}{{x}}}−\mathrm{1}−\sqrt{\mathrm{9}+\frac{\mathrm{3}}{{x}}}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{16}+\mathrm{4}{x}^{−\mathrm{1}} }−\mathrm{1}−\sqrt{\mathrm{9}+\mathrm{3}{x}^{−\mathrm{1}} }}{{x}^{−\mathrm{1}} } \\ $$$$\mathrm{applying}\:\mathrm{lhopital} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{16}+\mathrm{4}{x}^{−\mathrm{1}} \right)^{−\mathrm{1}/\mathrm{2}} ×\left(−\mathrm{4}{x}^{−\mathrm{2}} \right)−\left(\mathrm{9}+\mathrm{3}{x}^{−\mathrm{1}} \right)^{−\mathrm{1}/\mathrm{2}} ×\left(−\mathrm{3}{x}^{−\mathrm{2}} \right)}{−{x}^{−\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}}{\sqrt{\mathrm{16}+\mathrm{4}{x}^{−\mathrm{1}} }}−\frac{\mathrm{3}}{\sqrt{\mathrm{9}+\mathrm{3}{x}^{−\mathrm{1}} }} \\ $$$$=\:\mathrm{0} \\ $$
Commented by Joel577 last updated on 19/Sep/17
$${thank}\:{you}\:{very}\:{much} \\ $$