\(\lim\limits_{t \to +\infty} \sqrt{t^{2} + t} - t =\) ________
(Round off to one decimal place)
Answer
0.5
Explanation
Substituting directly gives the indeterminate form \(\infty - \infty\), so multiply and divide by the conjugate, then divide the numerator and denominator by \(t > 0\):
\[\sqrt{t^{2} + t} - t = \frac{(t^{2} + t) - t^{2}}{\sqrt{t^{2} + t} + t} = \frac{t}{\sqrt{t^{2} + t} + t} = \frac{1}{\sqrt{1 + \frac{1}{t}} + 1}.\]
As \(t \to +\infty\), \(\frac{1}{t} \to 0\), so the limit is \(\frac{1}{1 + 1} = 0.5\). As a check, the expression is about 0.4988 at \(t = 100\).
Answer: 0.5