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Question Number 90151 by JosephK last updated on 21/Apr/20

$$findthelimitof$$$$lim\frac{1}{t(\sqrt{1+t}}-\frac{1}{t}$$$$t\to 0$$

Commented by abdomathmax last updated on 21/Apr/20

$$f\left(t\right)=\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\Rightarrow f\left(t\right)=\frac{t-t\sqrt{1+t}}{t^2\sqrt{1+t}}$$$$=\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}wehave\sqrt{1+t}\sim 1+\frac{t}{2}+\frac{1}{2}\left(\frac{1}{2}\right)(\frac{1}{2}-1)t^2$$$$=1+\frac{t}{2}-\frac{t^2}{8}\Rightarrow 1-\sqrt{1+t}\sim -\frac{t}{2}+\frac{t^2}{8}$$$$t\sqrt{1+t}\sim t(1+\frac{t}{2}-\frac{t^2}{8})\sim t+\frac{t^2}{2}\Rightarrow$$$$f\left(t\right)\sim \frac{-\frac{t}{2}+\frac{t^2}{8}}{t+\frac{t^2}{2}}=\frac{-\frac{1}{2}+\frac{t}{8}}{1+\frac{t}{2}}\Rightarrow {lim}_{t\to 0}f\left(t\right)=-\frac{1}{2}$$

Answered by jagoll last updated on 22/Apr/20

$$\underset{\mathrm{t}\to 0}{\lim }\frac{1-\sqrt{1+\mathrm{t}}}{\mathrm{t}\sqrt{1+\mathrm{t}}}=\underset{\mathrm{t}\to 0}{\lim }\frac{1-(1+\frac{\mathrm{t}}{2})}{\mathrm{t}}$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{-\frac{\mathrm{t}}{2}}{\mathrm{t}}=-\frac{1}{2}$$

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