Question-32

\(\lim\limits _{t\rightarrow \infty } \ \sqrt{t^{2} +t} -t=\)____

(Round off to one decimal place)

NoteAnswer

\(0.5\)

NoteSolution

Rationalizing the numerator, we have:

\[ \begin{aligned} \sqrt{t^{2} +t} -t & =\frac{\left(\sqrt{t^{2} +t} -t\right)\left(\sqrt{t^{2} +t} +t\right)}{\sqrt{t^{2} +t} +t}\\ & \\ & =\frac{t^{2} +t-t^{2}}{\sqrt{t^{2} +t} +t}\\ & \\ & =\frac{t}{\sqrt{t^{2} +t} +t}\\ & \\ & =\frac{1}{\sqrt{1+\frac{1}{t}} +1} \end{aligned} \]

Now, we have

\[ \lim\limits _{t\rightarrow \infty } \ \frac{1}{\sqrt{1+\frac{1}{t}} +1} =\frac{1}{2} \]