Question Number 112855 by ruwedkabeh last updated on 10/Sep/20
$$ \\ $$$$\mathrm{1}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}{x}+\mathrm{4}}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}{\mathrm{4}{x}}=… \\ $$$$ \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}−\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} }−\sqrt{\mathrm{4}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}{x}}=… \\ $$$$ \\ $$
Answered by bobhans last updated on 10/Sep/20
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5x}+\mathrm{4}}−\sqrt{\mathrm{3x}+\mathrm{9}}}{\mathrm{4x}}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{5}}{\mathrm{x}}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }}−\sqrt{\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }}}{\mathrm{4}}\:=\:\mathrm{0} \\ $$
Answered by bobhans last updated on 10/Sep/20
$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}\left(\sqrt{\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{\mathrm{x}}+\mathrm{3}}\:−\sqrt{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{x}}+\mathrm{3}}\:\right)}{\mathrm{2x}}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{\mathrm{x}}+\mathrm{3}}−\sqrt{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$
Commented by ruwedkabeh last updated on 10/Sep/20
$${thank}\:{you}\:{very}\:{much}. \\ $$
Answered by mathmax by abdo last updated on 10/Sep/20
$$\left.\mathrm{1}\right)\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{5x}+\mathrm{4}}−\sqrt{\mathrm{3x}+\mathrm{9}}}{\mathrm{4x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{5x}}\sqrt{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{5x}}}−\sqrt{\mathrm{3x}}\sqrt{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{3x}}}}{\mathrm{4x}} \\ $$$$\sim\frac{\sqrt{\mathrm{5x}}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{5x}}\right)−\sqrt{\mathrm{3x}}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2x}}\right)}{\mathrm{4x}}\:=\frac{\left(\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{x}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5x}}}−\frac{\mathrm{3}\sqrt{\mathrm{3x}}}{\mathrm{2x}}}{\mathrm{4x}} \\ $$$$\sim\frac{\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{x}}}\:\rightarrow\mathrm{0}\:\:\left(\mathrm{x}\rightarrow+\infty\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{0} \\ $$