Answer
$$\frac{1}{2}$$
Work Step by Step
Since$$b_{n}=\sqrt{n^{2}+n}-\sqrt{n^{2}+1}$$
Then
\begin{align*}
\lim _{n \rightarrow \infty} b_{n}&=\lim _{n \rightarrow \infty} (\sqrt{n^{2}+n}-\sqrt{n^{2}+1})\frac{(\sqrt{n^{2}+n}+\sqrt{n^{2}+1})}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}}\\
&= \lim _{n \rightarrow \infty} \frac{\left(n^{2}+n\right)-\left(n^{2}+1\right)}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}}\\
&=\lim _{n \rightarrow \infty} \frac{n-1}{\sqrt{n^{2}+n}+\sqrt{n^{2}+1}} \\
&=\lim _{n \rightarrow \infty} \frac{\frac{n}{n}-\frac{1}{n}}{\sqrt{\frac{n^{2}}{n^{2}}+\frac{n}{n^{2}}}+\sqrt{\frac{n^{2}}{n^{2}}+\frac{1}{n^{2}}}}\\
&=\lim _{n \rightarrow \infty} \frac{1-\frac{1}{n}}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{1}{n^{2}}}}\\
&=\frac{1}{2}
\end{align*}
