Solutions

By multiplying $b_{n}$ by its conjugate, we obtain an easier form to deal with. That is,

$$b_{n} = n - \sqrt{n^2 + 2n} = n - \sqrt{n^2 + 2n} \cdot \frac{n + \sqrt{n^2 + 2n}}{n + \sqrt{n^2 + 2n}} = \frac{-2n}{n(1 + \sqrt{1 + \frac{2}{n}})} = \frac{-2}{1 + \sqrt{1 + \frac{2}{n}}}$$

Applying rule (i), we know that

$$\lim \frac{2}{n} = 0.$$

It then follows with rule (ii) that

$$\lim 1 + \frac{2}{n} = 1$$

and, by exercise 2.3.1 (b),

$$\lim \sqrt{1 + \frac{2}{n}} = \sqrt{1} = 1.$$

Finally, applying (ii) to

$$\lim 1 + \sqrt{1 + \frac{2}{n}}$$

and (iv) to $b_{n}$, we obtain

$$b_{n} = \frac{-2}{1 + \sqrt{1 + \frac{2}{n}}} = -\frac{2}{2} = -1.$$