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# Question-68171

Question Number 68171 by naka3546 last updated on 06/Sep/19
Commented by mathmax by abdo last updated on 06/Sep/19
$$\\$$$$\sqrt{\mathrm{5}−\mathrm{2}\sqrt{{x}}\:}\:{is}\:{not}\:{defined}\:{on}\:+\infty\:! \\$$
Commented by mathmax by abdo last updated on 07/Sep/19
$${i}\:{think}\:{the}\:{Q}\:{here}\:{is}\:{find}\:{lim}_{{x}\rightarrow+\infty} \frac{\sqrt{\mathrm{2}\sqrt{{x}}+\mathrm{5}}−\sqrt{\mathrm{2}\sqrt{{x}}−\mathrm{5}}}{\:\sqrt{{x}}−\mathrm{1}}={lim}_{{x}\rightarrow+\infty} {f}\left({x}\right) \\$$$${let}\:{treat}\:{this}\:\:{we}\:{use}\:{tbe}\:{changement}\:\sqrt{{x}}={t}\: \\$$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}_{{t}\rightarrow+\infty} \:\frac{\sqrt{\mathrm{2}{t}+\mathrm{5}}−\sqrt{\mathrm{2}{t}−\mathrm{5}}}{{t}−\mathrm{1}} \\$$$$\left.={lim}_{{t}\rightarrow+\infty} \:\frac{\mathrm{2}{t}+\mathrm{5}−\mathrm{2}{t}+\mathrm{5}}{\left({t}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}{t}+\mathrm{5}}+\sqrt{\mathrm{2}{t}−\mathrm{5}}\right)}\:={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{10}}{\left({t}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}{t}+\mathrm{5}}+\sqrt{\left.\mathrm{2}{t}−\mathrm{5}\right)}\right.}\right) \\$$$$=\mathrm{0} \\$$$$\\$$
Answered by Kunal12588 last updated on 06/Sep/19
$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\sqrt{\mathrm{5}+\mathrm{2}\sqrt{{x}}}−\sqrt{\mathrm{5}−\mathrm{2}\sqrt{{x}}}}{\:\sqrt{{x}}−\mathrm{1}} \\$$$$\sqrt{{x}}={t}\:;\:{x}\rightarrow\infty\:\Rightarrow\:{t}\rightarrow\infty \\$$$$\underset{{t}\rightarrow\infty} {{lim}}\frac{\sqrt{\mathrm{5}+\mathrm{2}{t}}−\sqrt{\mathrm{5}−\mathrm{2}{t}}}{{t}−\mathrm{1}} \\$$$$=\underset{{t}\rightarrow\infty} {{lim}}\frac{\sqrt{\frac{\mathrm{5}}{{t}^{\mathrm{2}} }+\frac{\mathrm{2}}{{t}}}−\sqrt{\frac{\mathrm{5}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}}}}{\mathrm{1}−\frac{\mathrm{1}}{{t}}} \\$$$$=\frac{\mathrm{0}}{\mathrm{1}}=\mathrm{0} \\$$