Question Number 90151 by JosephK last updated on 21/Apr/20
$${find}\:{the}\:{limit}\:{of}\: \\ $$$${lim}\:\:\frac{\mathrm{1}}{{t}\left(\sqrt{\mathrm{1}+{t}}\right.}−\frac{\mathrm{1}}{{t}} \\ $$$${t}\rightarrow\mathrm{0} \\ $$
Commented by abdomathmax last updated on 21/Apr/20
$${f}\left({t}\right)=\frac{\mathrm{1}}{{t}\sqrt{\mathrm{1}+{t}}}−\frac{\mathrm{1}}{{t}}\:\Rightarrow{f}\left({t}\right)=\frac{{t}−{t}\sqrt{\mathrm{1}+{t}}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}}} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{t}}}{{t}\sqrt{\mathrm{1}+{t}}}\:\:{we}\:{have}\:\sqrt{\mathrm{1}+{t}}\sim\:\mathrm{1}+\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right){t}^{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{{t}}{\mathrm{2}}−\frac{{t}^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\mathrm{1}−\sqrt{\mathrm{1}+{t}}\sim−\frac{{t}}{\mathrm{2}}+\frac{{t}^{\mathrm{2}} }{\mathrm{8}} \\ $$$${t}\sqrt{\mathrm{1}+{t}}\sim{t}\left(\mathrm{1}+\frac{{t}}{\mathrm{2}}−\frac{{t}^{\mathrm{2}} }{\mathrm{8}}\right)\sim{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({t}\right)\sim\frac{−\frac{{t}}{\mathrm{2}}+\frac{{t}^{\mathrm{2}} }{\mathrm{8}}}{{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}\:=\frac{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{t}}{\mathrm{8}}}{\mathrm{1}+\frac{{t}}{\mathrm{2}}}\:\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:\:{f}\left({t}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by jagoll last updated on 22/Apr/20
$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{t}}}{\mathrm{t}\sqrt{\mathrm{1}+\mathrm{t}}}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{t}} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\frac{\mathrm{t}}{\mathrm{2}}}{\mathrm{t}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$